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Mastering Algebra 
Advanced Level 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Book Title: Mastering Algebra - Advanced Level 
 
Author: Dan Hamilton 
 
Editor: Pat Eblen 
 
Cover design by: Kathleen Myers 
 
Copyright  2000 
All rights reserved. 
Printed in the United States of America. 
 
No part of this book may be reproduced, stored in a retrieval system, or transmitted in any form 
or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior 
written permission of the author. Request for permission or further information should be 
addressed to Hamilton Education Guides via info@hamiltoneducationguides.com. 
 
First published in 2000 
 
Library of Congress Catalog Card Number 99-97605 
Library of Congress Cataloging-in-Publication Data 
 
ISBN 0-9649954-3-3 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
This book is dedicated to my wife and children for their support and understanding. 
 
 
 
 
 
Hamilton Education Guides 
Book Series 
____________________________________________________________________________________ 
eBook and paperback versions available in the Amazon Kindle Store 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Hamilton Education Guides 
Manual Series 
____________________________________________________________________________________ 
eManual versions available in the Amazon Kindle Store 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Hamilton Education Guides 
Manual Series 
____________________________________________________________________________________ 
eManual versions available in the Amazon Kindle Store 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Hamilton Education Guides i 
General Contents 
 
 
Detailed Contents .................................................................................................................... iii 
 
Acknowledgments and Dedication ......................................................................................... vii 
 
Introduction and Overview ..................................................................................................... vii 
 
Chapter 1 Review of Introductory and Intermediate Algebra 
 Quick Reference to Chapter 1 Problems ...................................................................... 1 
 1.1 Exponents ......................................................................................................... 5 
 1.2 Radicals ............................................................................................................ 24 
 1.3 Factoring Polynomials ..................................................................................... 44 
 1.4 Quadratic Equations and Factoring .................................................................. 67 
 1.5 Algebraic Fractions .......................................................................................... 94 
 
Chapter 2 Functions of Real and Complex Variables 
 Quick Reference to Chapter 2 Problems ...................................................................... 115 
 2.1 Introduction to Functions of Real Variables .................................................... 117 
 2.2 Math Operations Involving Functions of Real Variables ................................ 124 
 2.3 Composite Functions of Real Variables .......................................................... 129 
 2.4 One-to-One and Inverse Functions of Real Variables ..................................... 139 
 2.5 Complex Numbers and Functions of Complex Variables ............................... 151 
 2.6 Math Operations Involving Complex Numbers ............................................... 159 
 
Chapter 3 Matrices 
 Quick Reference to Chapter 3 Problems ...................................................................... 167 
 3.1 Introduction to Matrices ................................................................................... 169 
 3.2 Matrix Operations ............................................................................................ 173 
 3.3 Determinants .................................................................................................... 185 
 3.4 Inverse Matrices ............................................................................................... 198 
 3.5 Solving Linear Systems.................................................................................... 210 
 
Chapter 4 Sequences and Series 
 Quick Reference to Chapter 4 Problems ...................................................................... 240 
 4.1 Sequences ......................................................................................................... 241 
 4.2 Series ................................................................................................................ 246 
 4.3 Arithmetic Sequences and Arithmetic Series................................................... 252 
 4.4 Geometric Sequences and Geometric Series ................................................... 259 
 4.5 Limits of Sequences and Series ....................................................................... 270 
 4.6 The Factorial Notation ..................................................................................... 281 
 
 
Mastering Algebra - Advanced Level General Contents 
Hamilton Education Guides ii 
Chapter 5 Differentiation 
 Quick Reference to Chapter 5 Problems ...................................................................... 292 
 5.1 The Difference Quotient Method ..................................................................... 293 
 5.2 Differentiation Rules Using the Prime Notation .............................................. 298 
 5.3 Differentiation Rules Using the d
dx
 Notation .................................................. 310 
 5.4 The Chain Rule ................................................................................................ 321 
 5.5 Implicit Differentiation .................................................................................... 336 
 5.6 The Derivative of Functions with Fractional Exponents ................................. 341 
 5.7 The Derivative of Radical Functions ..................................................................... 348 
 5.8 Higher Order Derivatives ................................................................................. 363 
 
Appendix Exercise Solutions 
 Chapter 1 Solutions...........................................................................................378 
 Chapter 2 Solutions...........................................................................................407 
 Chapter 3 Solutions...........................................................................................418 
 Chapter 4 Solutions...........................................................................................467 
 Chapter 5 Solutions...........................................................................................489 
 
Glossary ................................................................................................................................... 514 
 
Index ........................................................................................................................................522 
 
About the Author and Editor .................................................................................................. 529 
 
 
Hamilton Education Guides iii 
Detailed Contents 
Chapter 1 - Review of Introductory and Intermediate Algebra 
Quick Reference to Chapter 1 Problems 1 
1.1 Exponents 
1.1a Introduction to Integer Exponents ..................................................................................... 5 
Case I - Real Numbers Raised to Positive Integer Exponents 5 
Case II - Real Numbers Raised to Negative Integer Exponents 8 
1.1b Operations with Positive Integer Exponents ..................................................................... 10 
Case I - Multiplying Positive Integer Exponents 10 
Case II - Dividing Positive Integer Exponents 12 
Case III - Adding and Subtracting Positive Integer Exponents 14 
1.1c Operations with Negative Integer Exponents .................................................................... 16 
Case I - Multiplying Negative Integer Exponents 16 
Case II - Dividing Negative Integer Exponents 19 
Case III - Adding and Subtracting Negative Integer Exponents 21 
1.2 Radicals 
1.2a Introduction to Radicals .................................................................................................... 24 
Case I - Roots and Radical Expressions 24 
Case II - Rational, Irrational, Real, and Imaginary Numbers 26 
Case III - Simplifying Radical Expressions with Real Numbers as a Radicand 27 
1.2b Operations Involving Radical Expressions ....................................................................... 29 
Case I - Multiplying Monomial Expressions in Radical Form, with Real Numbers 29 
Case II - Multiplying Binomial Expressions in Radical Form, with Real Numbers 31 
Case III - Multiplying Monomial and Binomial Expressions in Radical Form, with Real 
Numbers 34 
Case IV - Rationalizing Radical Expressions - Monomial Denominators with Real Numbers 
 36 
Case V - Rationalizing Radical Expressions - Binomial Denominators with Real Numbers 
 39 
Case VI - Adding and Subtracting Radical Terms 42 
1.3 Factoring Polynomials 
1.3a Factoring Polynomials Using the Greatest Common Factoring Method .......................... 44 
Case I - Factoring the Greatest Common Factor to Monomial Terms 44 
Case II - Factoring the Greatest Common Factor to Binomial and Polynomial Terms 47 
1.3b Factoring Polynomials Using the Grouping Method ........................................................ 50 
1.3c Factoring Polynomials Using the Trial and Error Method ................................................ 52 
Case I - Factoring Trinomials of the Form ax bx c2 + + where a = 1 52 
Case II - Factoring Trinomials of the Form ax bx c2 + + where a 1 56 
1.3d Other Factoring Methods for Polynomials ........................................................................ 60 
Case I - Factoring Polynomials Using the Difference of Two Squares Method 60 
Case II - Factoring Polynomials Using the Sum and Difference of Two Cubes Method 63 
Case III - Factoring Perfect Square Trinomials 65 
 
Mastering Algebra - Advanced Level Detailed Contents 
Hamilton Education Guides iv 
1.4 Quadratic Equations and Factoring 
1.4a Quadratic Equations and the Quadratic Formula .............................................................. 67 
1.4b Solving Quadratic Equations Using the Quadratic Formula Method ............................... 69 
Case I - Solving Quadratic Equations of the Form ax bx c2 0+ + = where a = 1 69 
Case II - Solving Quadratic Equations of the Form ax bx c2 0+ + = where a 1 73 
1.4c Solving Quadratic Equations Using the Square Root Property Method ........................... 77 
1.4d Solving Quadratic Equations Using Completing-the-Square Method .............................. 81 
Case I - Solving Quadratic Equations of the Form ax bx c2 0+ + = , where a = 1 , by 
Completing the Square 81 
Case II - Solving Quadratic Equations of the Form ax bx c2 0+ + = , where a 1 , by 
Completing the Square 85 
1.4e How to Choose the Best Factoring or Solution Method ................................................... 89 
1.5 Algebraic Fractions 
1.5a Introduction to Algebraic Fractions .................................................................................. 94 
1.5b Simplifying Algebraic Fractions to Lower Terms ............................................................ 97 
1.5c Math Operations Involving Algebraic Fractions ............................................................... 99 
Case I - Addition and Subtraction of Algebraic Fractions with Common Denominators 99 
Case II - Addition and Subtraction of Algebraic Fractions without Common 
Denominators 101 
Case III - Multiplication of Algebraic Fractions 103 
Case IV - Division of Algebraic Fractions 105 
1.5d Math Operations Involving Complex Algebraic Fractions ............................................... 107 
Case I - Addition and Subtraction of Complex Algebraic Fractions 107 
Case II - Multiplication of Complex Algebraic Fractions 110 
Case III - Division of Complex Algebraic Fractions 112 
Chapter 2 - Functions of Real and Complex Variables 
Quick Reference to Chapter 2 Problems 115 
2.1 Introduction to Functions of Real Variables ............................................................ 117 
2.2 Math Operations Involving Functions of Real Variables ....................................... 124 
2.2.1 - Determining Odd and Even Functions 127 
2.3 Composite Functions of Real Variables .................................................................... 129 
Case I - Computing Composite Functions Using ( )xf and ( )xg 129 
Case II - Computing Composite Functions Using ( )xf , ( )xg , and ( )xh 134 
2.4 One-to-One and Inverse Functions of Real Variables ............................................. 139 
2.4.1 - One-to-One Functions 139 
Case I - The Function is Represented by a Set of Ordered Pairs 139 
Case II - The Function is Represented by an Equation 139 
2.4.2 - Inverse Functions 142 
Case I - The Function is Represented by a Set of Ordered Pairs 142 
Case II - The Function is Represented by an Equation 143 
Mastering Algebra - Advanced Level Detailed Contents 
Hamilton Education Guides v 
2.5 Complex Numbers and Functions of Complex Variables ....................................... 151 
2.6 Math Operations Involving Complex Numbers ....................................................... 159 
Case I - Addition and Subtraction of Complex Numbers 159 
Case II - Multiplication of Complex Numbers 160 
Case III - Division of Complex Numbers 161 
Case IV - Mixed Operations Involving Complex Numbers 163 
Chapter 3 - Matrices 
Quick Reference to Chapter 3 Problems 167 
3.1 Introduction to Matrices ............................................................................................ 169 
3.2 Matrix Operations 173 
Case I - Matrix Addition and Subtraction 173 
Case II - Matrix Multiplication 178 
3.3 Determinants ............................................................................................................... 185 
3.4 Inverse Matrices .......................................................................................................... 198 
3.5 Solving Linear Systems............................................................................................... 210 
Case I - Solving Linear Systems Using the Addition Method 210 
Case II - Solving Linear Systems Using the Substitution Method 214 
Case III - Solving Linear Systems Using the Inverse Matrices Method 218 
Case IV - Solving Linear Systems Using Cramer’s Rule 224 
Case V - Solving Linear Systems Using the Gaussian Elimination Method 228 
Case VI - Solving Linear Systems Using the Gauss-JordanElimination Method 234 
Chapter 4 - Sequences and Series 
Quick Reference to Chapter 4 Problems 240 
4.1 Sequences ..................................................................................................................... 241 
4.2 Series ............................................................................................................................ 246 
4.3 Arithmetic Sequences and Arithmetic Series ........................................................... 252 
4.4 Geometric Sequences and Geometric Series ............................................................. 259 
4.5 Limits of Sequences and Series .................................................................................. 270 
4.6 The Factorial Notation ............................................................................................... 281 
Chapter 5 - Differentiation 
Quick Reference to Chapter 5 Problems 292 
5.1 The Difference Quotient Method ............................................................................... 293 
5.2 Differentiation Rules Using the Prime Notation ...................................................... 298 
5.3 Differentiation Rules Using the d
dx
 Notation ........................................................... 310 
 
Mastering Algebra - Advanced Level Detailed Contents 
Hamilton Education Guides vi 
5.4 The Chain Rule ........................................................................................................... 321 
5.5 Implicit Differentiation .............................................................................................. 336 
5.6 The Derivative of Functions with Fractional Exponents......................................... 341 
5.7 The Derivative of Radical Functions ........................................................................ 348 
5.8 Higher Order Derivatives .......................................................................................... 363 
Appendix - Exercise Solutions 
Chapter 1 Solutions ............................................................................................................... 378 
Section 1.1a, Case I 378, Case II 378 
Section 1.1b, Case I 378, Case II 378, Case III 379 
Section 1.1c, Case I 379, Case II 380, Case III 380 
Section 1.2a, Case I 381, Case II 381, Case III 381 
Section 1.2b, Case I 382, Case II 382, Case III 383, Case IV 383, Case V 384, Case VI 385 
Section 1.3a, Case I 385, Case II 385 
Section 1.3b 386 
Section 1.3c, Case I 386, Case II 386 
Section 1.3d, Case I 387, Case II 387, Case III 387 
Section 1.4a 387 
Section 1.4b, Case I 388, Case II 390 
Section 1.4c 392 
Section 1.4d, Case I 394, Case II 396 
Section 1.4e 399 
Section 1.5a 401 
Section 1.5b 402 
Section 1.5c, Case I 402, Case II 402, Case III 403, Case IV 403 
Section 1.5d, Case I 404, Case II 405, Case III 405 
Chapter 2 Solutions ............................................................................................................... 407 
Section 2.1 407 Section 2.2 409 Section 2.3 411 
Section 2.4 412 Section 2.5 413 
Section 2.6, Case I 415, Case II 415, Case III 416, Case IV 416 
Chapter 3 Solutions ............................................................................................................... 418 
Section 3.1 418 
Section 3.2, Case I 419, Case II 422 
Section 3.3 427 
Section 3.4 430 
Section 3.5, Case I 438, Cases II 440, Case III 443, Case IV 449, Case V 456, 
 Case VI 461 
Chapter 4 Solutions ............................................................................................................... 467 
Section 4.1 467 Section 4.2 470 Section 4.3 472 
Section 4.4 476 Section 4.5 480 Section 4.6 484 
Chapter 5 Solutions ............................................................................................................... 489 
Section 5.1 489 Section 5.2 491 Section 5.3 495 
Section 5.4 500 Section 5.5 505 Section 5.6 506 
Section 5.7 508 Section 5.8 511 
Hamilton Education Guides vii 
Acknowledgments and Dedication 
 
 I would like to acknowledge my wife and children for giving me inspiration and for their 
understanding and patience in allowing me to take on the task of writing this book. I am grateful 
to Pat Eblen for his editorial comments. As always, his constructive comments and suggestions 
on clearer presentation of topics truly elevated the usefulness of this book. His continual support 
and recommendations to further enhance this book is greatly appreciated. I would also like to 
acknowledge and give my thanks to many education professionals who provided comments to 
further enhance this book. Finally, my special thanks to Kathleen Myers for her outstanding 
cover design. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Hamilton Education Guides viii 
 Introduction and Overview 
 
As we start the new century, with anticipation and a great deal of aspiration, a primary 
hope of our nation is that the new millennium will bring new discoveries and advancements to the 
fields of science, technology, communication, medicine, and space exploration - to mention a few 
- that will ultimately enhance human life on this planet. This hope and aspiration can only be 
fulfilled if the individuals in our society obtain a thorough knowledge of technical concepts. A 
key to achieving this goal is by providing quality educational materials that can build and develop 
individuals abilities in solving complex problems. 
Similar to the previous books published by the Hamilton Education Guides, the intent of 
this book is to build a strong foundation by increasing student confidence in solving mathematical 
problems. To achieve this objective, the author has diligently tried to address each subject in a 
clear, concise, and easy to understand step-by-step format. A great deal of effort has been made to 
ensure the subjects presented in each chapter are explained simply, thoroughly, and adequately. It 
is the authors hope that this book can fulfill these objectives by building a solid foundation in 
pursuit of more advanced technical concepts. 
 The scope of this book is intended for educational levels ranging from the 11th grade to 
adult. The book can also be used by students in home study programs, parents, teachers, special 
education programs, tutors, high schools, preparatory schools, and adult educational programs 
including colleges and universities as a main text, a thorough reference, or a supplementary book. 
A thorough knowledge of algebraic concepts in subject areas such as linear equations and 
inequalities, fractional operations, exponents, radicals, polynomials, factorization, and quadratic 
equations is required. 
 “Mastering Algebra: Advanced Level” is the third and final book in a series on algebra. It 
addresses subjects such as functions of real and complex variables, matrices, sequences, series, 
limits, factorials, and differentiation. In the second book “Mastering Algebra: Intermediate 
Level” students are introduced to subjects such as linear and non linear equations, inequalities, 
factoring, quadratic equations, algebraic fractions, and logarithms. In the first book “Mastering 
Algebra: An Introduction” topics such as integer fractions, exponents, radicals, fractional 
exponents, scientific notations, and polynomials are introduced. 
 “Mastering Algebra: Advanced Level” is divided into five chapters. Chapter 1 reviews, in 
some detail, selected subjects addressed in the “Mastering Algebra: An Introduction” and the 
“Mastering Algebra: Intermediate Level” books. Topics such as exponents, radicals, factoring 
polynomials, quadratic equations andfactoring, and operations involving with real and complex 
algebraic fractions are reviewed. Functions of real and complex variables are addressed in Chapter 
2. In this chapter students learn how to identify and solve various math operations involving 
functions. In addition, the steps as to how composite and inverse functions are computed are also 
addressed in this chapter. Finally, math operations involving addition, subtraction, multiplication, 
and division of complex numbers are discussed in Chapter 2. Matrix operations including the steps 
for computing the determinant and inverse of a matrix are discussed in Chapter 3. Solutions to 
linear systems using various methods such as the Addition, the Substitution, the Inverse Matrices, 
the Cramer’s Rule, the Gaussian Elimination, and the Gauss-Jordan Elimination method are 
addressed in Chapter 3. Students are encouraged to gain a thorough understanding of the various 
Mastering Algebra – Advanced Level Introduction and Overview 
Hamilton Education Guides ix 
solution methods for solving linear systems of equations. Sequences and series are introduced in 
Chapter 4. Additionally, how to compute and find the limit of arithmetic and geometric sequences 
and series including expansion and simplification of factorial expressions are also discussed in this 
chapter. Derivatives and its applicable differentiation rules using the Prime and ddx notations are 
introduced in Chapter 5. Use of the Chain rule in solving different types of equations is also 
discussed in this chapter. Additionally, the steps for solving higher order equations including the 
implicit differentiation method is introduced in Chapter 5. Finally, detailed solutions to the 
exercises are provided in the Appendix. Students are encouraged to solve each problem in the same 
detail and step-by-step format as shown in the text. 
 In keeping with our commitment of excellence in providing clear, easy to follow, and 
concise educational materials to our readers, I believe this book will again add value to the Hamilton 
Education Guides series for its clarity and special attention to detail. I hope readers of this book 
will find it valuable as a learning tool and a reference. Any comments or suggestions for 
improvement of this book will be appreciated. 
 
With best wishes, 
 
Dan Hamilton 
 
Mastering Algebra - Advanced Level Quick Reference to Chapter 1 Problems 
Hamilton Education Guides 1 
Chapter 1 
Review of Introductory and Intermediate Algebra 
 
Quick Reference to Chapter 1 Problems 
1.1 Exponents 
1.1a Introduction to Integer Exponents .................................................................................. 5 
Case I - Real Numbers Raised to Positive Integer Exponents, p. 5 
23 = ; ( )−3 5 = ; − 83 = 
Case II - Real Numbers Raised to Negative Integer Exponents, p. 8 
4 3− = ; ( )− −8 3 = ; − −2 4 = 
1.1b Operations with Positive Integer Exponents .................................................................. 10 
Case I - Multiplying Positive Integer Exponents, p. 10 
( ) ( )x y x y y3 2 2 3⋅ ⋅ = ; ( )e e e e3 5 2432⋅ −



 = ; ( ) ( ) ( )2 3 5 22 3 2− ⋅ ⋅ −x y xy = 
Case II - Dividing Positive Integer Exponents, p. 12 
2
4 3 4
ab
a b−
 = ; u v w
u v
u
v
2 3 2
7 5 28





 ⋅




 = ; 4
3 2
2 3
k lm
kl m
 = 
Case III - Adding and Subtracting Positive Integer Exponents, p. 14 
 x y x y3 2 3 23 2 5+ + − + = ; ( ) ( )a a a3 2 3 32 4 4 20+ + − + = ; a a a ab b b b3 2 3 22 4 5 3+ − + + = 
1.1c Operations with Negative Integer Exponents ................................................................ 16 
Case I - Multiplying Negative Integer Exponents, p. 16 
5 52 3 3 1− − − −⋅ ⋅ ⋅ ⋅ ⋅a b a b = ; ( ) ( )x y z x y z− − − −⋅2 2 2 1 3 4 = ; ( ) ( )3 2 3 2 22 1 1 2− − − −⋅ ⋅ ⋅ ⋅ = 
Case II - Dividing Negative Integer Exponents, p. 19 
2
2
3 1 3
1 2
− − −
− −
u v
u v
 = ; e f e
f e
− −
− −
5 2 0
3 5 42
 = ; 
( )
2
2
3 1
2 3
− −
− −
⋅
−
a
a
 = 
Case III - Adding and Subtracting Negative Integer Exponents, p. 21 
 x x x x− − − − −+ + − +1 2 1 2 22 4 5 = ; a b a b− − − −− + +1 1 1 12 3 = ; k k kn n n− − − −+ − +2 3 2 23 2 = 
1.2 Radicals 
1.2a Introduction to Radicals ................................................................................................. 24 
Case I - Roots and Radical Expressions, p. 24 
64 = ; 3753 = ; − 3242 = 
Case II - Rational, Irrational, Real, and Imaginary Numbers, p. 26 
Mastering Algebra - Advanced Level Quick Reference to Chapter 1 Problems 
Hamilton Education Guides 2 
25 is a rational number ; 7 is an irrational number ; − −3 is not a real number 
Case III - Simplifying Radical Expressions with Real Numbers as a Radicand, p. 27 
−
−
3
2
400 = ; 162
9
3
 = ; 1
5
10002
−
 = 
1.2b Operations Involving Radical Expressions .................................................................... 29 
Case I - Multiplying Monomial Expressions in Radical Form, with Real Numbers, p. 29 
5 15⋅ = ; ( ) ( )− ⋅ −2 512 5 1083 3 = ; 6 48 45⋅ ⋅ = 
Case II - Multiplying Binomial Expressions in Radical Form, with Real Numbers, p. 31 
( ) ( )2 2 5 8+ ⋅ − = ; ( ) ( )2 162 3 3 2 54 4+ ⋅ + = ; ( ) ( )24 3 60 25 72+ ⋅ − = 
Case III - Multiplying Monomial and Binomial Expressions in Radical Form, with Real 
 Numbers, p. 34 
( )5 50 2 27⋅ + = ; ( )− ⋅ −2 24 36 125 = ; ( )− ⋅ −2 4 64 1624 4 4 = 
Case IV - Rationalizing Radical Expressions - Monomial Denominators with Real 
 Numbers, p. 36 
−8 3
32 45
 = ; 
( )5 2 3
7
2− ⋅
−
 = ; 3 8
81
5
5 = 
Case V - Rationalizing Radical Expressions - Binomial Denominators with Real 
 Numbers, p. 39 
8
2 2−
 = ; 125
3 5−
 = ; 3 5
3 5
+
+
 = 
Case VI - Adding and Subtracting Radical Terms, p. 42 
6 2 4 2+ = ; 20 3 8 3 5 35 5 5− + = ; a xy b xy c xy d+ − −3 2 = 
1.3 Factoring Polynomials 
1.3a Factoring Polynomials Using the Greatest Common Factoring Method ........... 44 
Case I - Factoring the Greatest Common Factor to Monomial Terms, p. 44 
Find the G.C.F. to: 6 82 3x and x ; 8 242 3 3x yz and xy z ; r s rs and r s3 3 2 28 9, , 
Case II - Factoring the Greatest Common Factor to Binomial and Polynomial Terms, p. 47 
6 23 2 2 2 2a b c a bc− ; 12 363 2 2 2x y z x z+ ; 8 4 22 3 2 2a b ab a b+ − 
1.3b Factoring Polynomials Using the Grouping Method ......................................... 50 
x x x2 4 9 36− − + = ; 60 24 15 62m m m+ − − = ; ( )5 15 152x y x y+ + + = 
1.3c Factoring Polynomials Using the Trial and Error Method ................................. 52 
Case I - Factoring Trinomials of the Form ax bx c2 + + where a = 1 , p. 52 
Mastering Algebra - Advanced Level Quick Reference to Chapter 1 Problems 
Hamilton Education Guides 3 
x x2 16 55− + = ; x x2 2 48+ − = ; x x2 6 40− − = 
Case II - Factoring Trinomials of the Form ax bx c2 + + where a 1 , p. 56 
6 23 202x x+ + = ; 10 9 912x x− − = ; 2 19 352 2x xy y− + = 
1.3d Other Factoring Methods for Polynomials......................................................... 60 
Case I - Factoring Polynomials Using the Difference of Two Squares Method, p. 60 
x y x y4 2 3− = ; 81 4 4m n− = ; ( )u v2 21− + = 
Case II - Factoring Polynomials Using the Sum and Difference of Two Cubes Method, p. 63 
x y3 3 27− = ; ( )x y− +1 3 3 = ; ( )x y3 35− + = 
Case III - Factoring Perfect Square Trinomials, p. 65 
16 24 92 2x xy y+ + = ; 25 64 802 2r s rs+ − = ; 9 42 494 2 2 4x x y y− + = 
1.4 Quadratic Equations and Factoring 
1.4a Quadratic Equations and the Quadratic Formula ............................................... 67 
ax bx c2 0+ + = ; x b b ac
a
=
− ± −2 4
2
 
1.4b Solving Quadratic Equations Using the Quadratic Formula Method ................69 
Case I - Solving Quadratic Equations of the Form ax bx c2 0+ + = where a = 1 , p. 69 
x x2 5 4+ = − ; x x2 12 35= − − ; x x2 5 6 0− + = 
Case II - Solving Quadratic Equations of the Form ax bx c2 0+ + = where a 1 , p. 73 
2 5 32x x+ = − ; 15 7 22x x= − + ; 4 4 32 2x xy y+ = 
1.4c Solving Quadratic Equations Using the Square Root Property Method ............ 77 
( )x + =4 362 ; ( )x − =2 252 ; ( )2 4 162x − = 
1.4d Solving Quadratic Equations Using Completing-the-Square Method ............... 81 
Case I - Solving Quadratic Equations of the Form ax bx c2 0+ + = , where a = 1 , by 
Completing the Square, p. 81 
x x2 8 5 0+ + = ; x x2 4 3 0− + = ; x x2 6 0+ − = 
Case II - Solving Quadratic Equations of the Form ax bx c2 0+ + = , where a 1 , by 
Completing the Square, p. 85 
3 16 5 02x x− + = ; 2 3 6 02x x+ − = ; 3 12 4 02t t+ − = 
1.4e How to Choose the Best Factoring or Solution Method .................................... 89 
 Use different methods to solve: x2 25= ; x x2 11 24 0+ + = ; x x2 5 2 0+ + = 
 
Mastering Algebra - Advanced Level Quick Reference to Chapter 1 Problems 
Hamilton Education Guides 4 
1.5 Algebraic Fractions 
1.5a Introduction to Algebraic Fractions ................................................................... 94 
−
−
+
= − −



= +
a
b
a
b
a
b
 ; 2
1
1
x
is not defined when x
+
= − ; 1 2 2
2
+
=
+
=
+x
x
x
x
a ax
ax
 
1.5b Simplifying Algebraic Fractions to Lower Terms ............................................. 97 
3
15
2 2
3 2
a b c
ab c
 = ; 3 6
62
x
x x
+
− −
 = ; x x
x x x
3
3 22 3
−
− −
 = 
1.5c Math Operations Involving Algebraic Fractions ................................................ 99 
Case I - Addition and Subtraction of Algebraic Fractions with Common Denominators, 
 , p. 99 
5 3x
x y
x
x y+
−
+
 = ; 3
2
2
2
2
22 3 2 3 2 3
a b
a b
a b
a b
a b
a b
+
+
−
+
− = ; ( )( ) ( )( )
y y
y y
y y
y y
3 33
2 3
2 3
2 3
+
− +
−
+ −
− +
 = 
Case II - Addition and Subtraction of Algebraic Fractions without Common Denominators 
 , p. 101 
2
2
4
3x x−
+
+
 = ; x
yz
y
xz
z
xy
+ + = ; y
x xy
x
xy y2 2−
−
−
 = 
Case III - Multiplication of Algebraic Fractions, p. 103 
x y
x
y
x
2
3⋅ = ; u v w uv
w
v w
2 2
2 2 3
1
⋅ ⋅ = ; ab
a b
a
b
2
2
3
2
1 4
⋅ ⋅





 = 
Case IV - Division of Algebraic Fractions, p. 105 
1 3
3 3 2 2x y z
z
x y
÷ = ; a b ab
ab
2 2
33
÷ = ; 1 32
2
2xy
x
y
y
x
÷





 ÷ = 
1.5d Math Operations Involving Complex Algebraic Fractions ................................ 107 
Case I - Addition and Subtraction of Complex Algebraic Fractions, p. 107 
1
2
2
3
3
3
5
2
x y
y x
+
−
 = ; 
3 1
1
3 2
2
2
x y
xy
x y
xy
−
+
 = ; 
1 1
1 1
a b
a b
a
b
+
−
− = 
Case II - Multiplication of Complex Algebraic Fractions, p. 110 
xy
x
x
x y
xy
2
2 2 1
⋅
⋅
 = ; 
x y z
z
z
x y
xy
2 2
2 3 3
1
⋅
 = ; 
xyz
xy
x
y
x
y
y
x
2 2
3
2
3
⋅
⋅
 = 
Case III - Division of Complex Algebraic Fractions, p. 112 
a b c
b c
a
ab
a
b
3 2
2 3
3
3
2
÷
÷
 = ; 
uv w
w
u v
u
u
v
2
3
2 3
÷
 = ; 3
2
2
2
xyz
xy
z
z
xy
xyz
÷
÷
 = 
Hamilton Education Guides 5 
Chapter 1 – Review of Introductory and 
Intermediate Algebra 
 
The objective of this chapter is to review in some detail selected subjects that were addressed in 
the “Mastering Algebra - An Introduction” and “Mastering Algebra - Intermediate Level” books. 
Students are expected to be familiar with the subjects presented in these books (Parentheses and 
Brackets, Integer Fractions, Exponents, Radicals, Fractional Exponents, Scientific Notation, 
Polynomials, One Variable Linear Equations and Inequalities, Factoring Polynomials, Quadratic 
Equations and Factoring, Algebraic Fractions, and Logarithms) before proceeding with the 
subjects introduced in this book. Students who may need to review additional examples, beyond 
what is shown in this chapter, need to study the “Mastering Algebra - An Introduction“ and 
“Mastering Algebra - Intermediate Level” books. 
Real numbers raised to positive and negative integer exponents are reviewed in Section 1.1. 
Mathematical operations involving addition, subtraction, multiplication, and division of positive 
and negative integer exponents is also discussed in Section 1.1. An introduction to radical 
expressions, the steps in simplifying radicals, as well as how rational, irrational, real, and 
imaginary numbers are identified is reviewed in Section 1.2. In addition, multiplication, 
division, addition, and subtraction of radical expressions is addressed in this section. Different 
methods for factoring polynomials and quadratic equations are reviewed in Sections 1.3 and 1.4, 
respectively. Algebraic fractions and their simplification to lower terms is addressed in Section 
1.5. Addition, subtraction, multiplication, and division of simple and complex algebraic 
fractions are also reviewed in this section. Additional examples, followed by practice problems, 
are provided in the cases presented in each section to help meet the objective of this chapter. 
1.1 Exponents 
In this section an introduction to exponents and how real numbers are raised to positive or 
negative integer exponents are reviewed in section 1.1a, Cases I and II. In addition, operations 
involving positive and negative integer exponents are addressed in sections 1.1b, Cases I through 
III, and 1.1c, Cases I through III, respectively. 
 
1.1a Introduction to Integer Exponents 
Integer exponents are defined as a n where a is referred to as the base, and n is the integer 
exponent. Note that the base a can be a real number or a variable. The integer exponent n can 
be a positive or a negative integer. Real numbers raised to positive and negative integer 
exponents (Cases I and II) are addressed below: 
Case I Real Numbers Raised to Positive Integer Exponents 
In general, real numbers raised to positive integer exponents are shown as: 
a n+ = a n = a a a a a⋅ ⋅ ⋅ ... where n is a positive eger and aint ≠ 0 
For example, 
8 4+ = 84 = 8 8 8 8⋅ ⋅ ⋅ = 4096 
Real numbers raised to a positive integer exponent are solved using the following steps: 
Mastering Algebra - Advanced Level 1.1a Introduction to Integer Exponents 
Hamilton Education Guides 6 
Step 1 Multiply the base a by itself as many times as the number specified in the exponent. 
For example, 25 implies that multiply 2 by itself 5 times, i.e., 2 2 2 2 2 25 = ⋅ ⋅ ⋅ ⋅ . 
Step 2 Multiply the real numbers to obtain the product, i.e., 2 2 2 2 2 32⋅ ⋅ ⋅ ⋅ = . 
Examples with Steps 
The following examples show the steps as to how real numbers raised to positive integer 
exponents are solved: 
Example 1.1-1 
 23 = 
Solution: 
 Step 1 23 = 2 2 2⋅ ⋅ 
 
 Step 2 2 2 2⋅ ⋅ = 8 
Example 1.1-2 
 124. = 
Solution: 
 Step 1 124. = ( ) ( ) ( ) ( )12 12 12 12. . . .⋅ ⋅ ⋅ 
 
 Step 2 ( ) ( ) ( ) ( )12 12 12 12. . . .⋅ ⋅ ⋅ = 2 074. 
Example 1.1-3 
 ( )−3 5 = 
Solution: 
 Step 1 ( )−3 5 = ( ) ( ) ( ) ( ) ( )− ⋅ − ⋅ − ⋅ − ⋅ −3 3 3 3 3 
 
 Step 2 ( ) ( ) ( ) ( ) ( )− ⋅ − ⋅ − ⋅ − ⋅ −3 3 3 3 3 = −243 
Note that: 
• A negative number raised to an even integer exponent such as 2 , 4 , 6 , 8 , 10 , 12 , etc. is 
always positive. For example, 
( )−3 6 = ( )+3 6 = +729 = 729 ( )−2 2 = ( )+2 2 = +4 = 4 ( )−5 4 = ( )+5 4 = +625 = 625 
• A negative number raised to an odd integer exponent such as 1, 3 , 5 , 7 , 9 , 11, etc. is 
always negative. For example, 
( )−3 5 = −243 ( )−2 3 = −8 ( )−3 7 = −2187 
Additional Examples - Real Numbers Raised to Positive Integer Exponents 
The following examples further illustrate how to solve real numbers raised to positive integer 
exponents: 
 
Mastering Algebra - Advanced Level 1.1a Introductionto Integer Exponents 
Hamilton Education Guides 7 
Example 1.1-4 
( )−10 0 = 1 (See the notes on page 9 relative to numbers raised to the zero power.) 
Example 1.1-5 
( )− 6 5 = ( )− ⋅ ⋅ ⋅ ⋅6 6 6 6 6 = ( )− 7776 = −7776 
Example 1.1-6 
( )−4 25 3. = ( ) ( ) ( )− ⋅ − ⋅ −4 25 4 25 4 25. . . = −76 77. 
Example 1.1-7 
( )10 45 4. = ( ) ( ) ( ) ( )10 45 10 45 10 45 10 45. . . .⋅ ⋅ ⋅ = 11925 19. 
Example 1.1-8 
( )− −20 3 = ( ) ( ) ( )[ ]− − ⋅ − ⋅ −20 20 20 = [ ]− −8000 = +8000 = 8000 
 
Practice Problems - Real Numbers Raised to Positive Integer Exponents 
Section 1.1a Case I Practice Problems - Solve the following exponential expressions with real 
numbers raised to positive integer exponents: 
1. 43 = 
 
2. ( )−10 4 = 3. 0 253. = 
4. 125 = 
 
5. ( )− 3 5 = 6. 4890 = 
 
Mastering Algebra - Advanced Level 1.1a Introduction To Integer Exponents 
Hamilton Education Guides 8 
Case II Real Numbers Raised to Negative Integer Exponents 
Negative integer exponents are defined as a n− where a is referred to as the base, and n is the 
integer exponent. Again, note that the base a can be a real number or a variable. The integer 
exponent n can be a positive or a negative integer. In this section, real numbers raised to 
negative integer exponents are addressed. 
In general, real numbers raised to negative integer exponents are shown as: 
a n− = 1
a n+
 = 1
a n
 = 1
a a a a a⋅ ⋅ ⋅ ⋅...
 where n is a positive eger and aint ≠ 0 
For example, 
5 4− = 1
54
 = 1
5 5 5 5⋅ ⋅ ⋅
 = 1
625
 
Real numbers raised to a negative integer exponent are solved using the following steps: 
Step 1 Change the negative integer exponent a n− to a positive integer exponent of the form 
1
an
. For example, change 3 4− to 1
34
. 
Step 2 Multiply the base a in the denominator by itself as many times as the number 
specified in the exponent. For example, rewrite 1
34
 as 1
3 3 3 3⋅ ⋅ ⋅
. 
Step 3 Multiply the real numbers in the denominator to obtain the answer, i.e., 1
3 3 3 3
1
81⋅ ⋅ ⋅
= . 
Examples with Steps 
The following examples show the steps as to how real numbers raised to negative integer 
exponents are solved: 
Example 1.1-9 
 4 3− = 
Solution: 
 Step 1 4 3− = 1
43
 
 
 Step 2 1
43
 = 1
4 4 4⋅ ⋅
 
 
 Step 3 1
4 4 4⋅ ⋅
 = 164 
Example 1.1-10 
 ( )− −8 3 = 
Solution: 
 Step 1 ( )− −8 3 = 
( )
1
8 3−
 
 Step 2 
( )
1
8 3−
 = ( ) ( ) ( )
1
8 8 8− ⋅ − ⋅ −
 
Mastering Algebra - Advanced Level 1.1a Introduction To Integer Exponents 
Hamilton Education Guides 9 
 Step 3 ( ) ( ) ( )
1
8 8 8− ⋅ − ⋅ −
 = 1
512−
 = − 1512 
Additional Examples - Real Numbers Raised to Negative Integer Exponents 
The following examples further illustrate how to solve real numbers raised to negative integer 
exponents: 
Example 1.1-11 
2 3− = 1
23
 = 1
2 2 2⋅ ⋅
 = 18 
Example 1.1-12 
( )− −6 4 = 
( )
−
1
6 4
 = − 1
64
 = −
⋅ ⋅ ⋅
1
6 6 6 6
 = − 11296 
Example 1.1-13 
( )5 2 4. − = 
( )
1
5 2 4.
 = 1
5 24.
 = ( ) ( ) ( ) ( )
1
5 2 5 2 5 2 5 2. . . .⋅ ⋅ ⋅
 = 173116. 
Example 1.1-14 
( )− −9 4 = 
( )
1
9 4−
 = ( ) ( ) ( ) ( )
1
9 9 9 9− ⋅ − ⋅ − ⋅ −
 = 1
6561+
 = 16561 
Example 1.1-15 
( )− − −4 5 3. = 
( )
−
−
1
4 5 3.
 = ( ) ( ) ( )
−
− ⋅ − ⋅ −
1
4 5 4 5 4 5. . .
 = −
−
1
91125.
 = + 1
91125.
 = 191125. 
Note 1: Any number or variable raised to the zero power is always equal to 1. For example, 
 55 10 = , ( )− =15 10 , ( )5 689 763 10, , = , ( )[ ]5 2 8 10x + − = , ( )[ ]axy 3 0 1= , and x y z3 2 0 1  = 
Note 2: Zero raised to the zero power is not defined, i.e., 00 is undefined. 
Note 3: Any number or variable divided by zero is not defined, i.e., 1
0
, x
0
, 355
0
, etc. are undefined. 
Note 4: Zero divided by any number or variable is always equal to zero. For example, 0
1
0= , 
 0
2 560
0
,
= , 0
10
0= , 0 02x
= , and 0 0
23 a
= . 
Practice Problems - Real Numbers Raised to Negative Integer Exponents 
Section 1.1a Case II Practice Problems - Solve the following exponential expressions with real 
numbers raised to negative integer exponents: 
1. 4 3− = 
 
2. ( )− −5 4 = 3. 0 25 3. − = 
4. 12 5− = 
 
5. ( )− −3 4 = 6. 48 2− = 
 
Mastering Algebra - Advanced Level 1.1b Operations with Positive Integer Exponents 
Hamilton Education Guides 10 
1.1b Operations with Positive Integer Exponents 
To multiply, divide, add, and subtract integer exponents, we need to know the following laws of 
exponents (shown in Table 1.1-1). These laws are used to simplify the work in solving problems 
with exponential expressions and should be memorized. 
Table 1.1-1: Exponent Laws 1 through 7 (Positive Integer Exponents) 
I. Multiplication a a am n m n⋅ = + When multiplying positive exponential terms, if 
 bases a are the same, add the exponents m and 
 n . 
II. Power of a Power ( )a am n mn= When raising an exponential term to a power, 
 multiply the powers (exponents) m and n . 
III. Power of a Product ( )a b a bm m m⋅ = ⋅ When raising a product to a power, raise each factor 
 a and b to the power m . 
IV. Power of a Fraction a
b
a
b
m m
m




= When raising a fraction to a power, raise the 
 numerator and the denominator to the power m . 
V. Division a
a
a a a
m
n
m n m n= ⋅ =− − When dividing exponential terms, if the bases a 
 are the same, subtract exponents m and n . 
VI. Negative Power a
a
n
n
− =
1 A non-zero base a raised to the −n power equals 
 1 divided by the base a to the n power. 
VII. Zero Power a0 1= A non-zero base a raised to the zero power is 
 always equal to 1. 
 
In this section students learn how to multiply (Case I), divide (Case II), and add or subtract (Case 
III) positive integer exponents by one another. 
Case I Multiplying Positive Integer Exponents 
Positive integer exponents are multiplied by each other using the following steps and the exponent 
laws I through III shown in Table 1.1-1. 
Step 1 Group the exponential terms with similar bases. 
Step 2 Apply the Multiplication Law (Law I) from Table 1.1-1 and simplify the exponential 
expressions by adding the exponents with similar bases. 
Examples with Steps 
The following examples show the steps as to how positive integer exponents are multiplied by 
one another: 
Example 1.1-16 
 ( ) ( )x y x y y3 2 2 3⋅ ⋅ = 
 
Mastering Algebra - Advanced Level 1.1b Operations with Positive Integer Exponents 
Hamilton Education Guides 11 
Solution: 
 Step 1 ( ) ( )x y x y y3 2 2 3⋅ ⋅ = ( ) ( )x x y y y3 2 3 2⋅ = ( ) ( )x x y y y3 2 3 2 1⋅ 
 
 Step 2 ( ) ( )x x y y y3 2 3 2 1⋅ = ( ) ( )x y3 2 3 2 1+ + +⋅ = x y5 6⋅ = x y5 6 
Example 1.1-17 
 ( )−


⋅ ⋅ −



1
5
10 1
4
2 2a a b a b = 
Solution: 
 Step 1 ( )−


⋅ ⋅ −



1
5
10 1
4
2 2a ab ab = ( ) ( )− × − ×  ⋅ ⋅
1
5
1
4
10 2 2a aa b b = ( ) ( )1020 2 1 1 2 1




⋅ ⋅a a a b b 
 
 Step 2 ( ) ( )1020 2 1 1 2 1




⋅ ⋅a a a b b = ( ) ( )/ // / ⋅ ⋅+ + +
10
20
2
2 1 1 2 1a b = 1
2
4 3⋅ ⋅a b = 12
4 3a b 
Note - Non zero numbers or variables raised to the zero power are always equal to 1, i.e., 10 10 = , 
( )23456 10 = , a for a0 1 0= ≠ , ( )a b for a b⋅ = ⋅ ≠0 1 0 , ( )x y z for x y z⋅ ⋅ = ⋅ ⋅ ≠0 1 0 , etc. 
Additional Examples - Multiplying Positive Integer Exponents 
The following examples further illustrate how positive exponential terms are multiplied: 
Example 1.1-18 
2 23 5⋅ = 23 5+ = 28 = 256 
Example 1.1-19 
3 3 3 30 2 3⋅ ⋅ ⋅ = 3 3 3 30 1 2 3⋅ ⋅ ⋅ = 30 1 2 3+ + + = 36 = 729 
Example 1.1-20 
x y z x y z x2 2 3 2 2 4⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = ( ) ( ) ( )x x x y y z z2 2 2 2 3 4⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = ( ) ( ) ( )x y z2 2 1 2 2 3 4+ + + +⋅ ⋅ = x y z5 4 7 
Example 1.1-21 
( ) ( ) ( )− − ⋅ − ⋅ −3 2 4 52 2 2k p k p = ( ) ( )[ ] ( ) ( )− ⋅ − × − ⋅ ⋅5 4 52 2 2k k p p = ( ) ( )[] ( ) ( )+ ⋅ + ⋅ ⋅25 20 2 1 2 1k k p p 
 
= ( ) ( )500 2 1 2 1⋅ ⋅+ +k p = 500 3 3k p 
Practice Problems - Multiplying Positive Integer Exponents 
Section 1.1b Case I Practice Problems - Multiply the following positive integer exponents: 
1. x x x2 3⋅ ⋅ = 
 
2. 2 2 0 3 2⋅ ⋅ ⋅ ⋅a b a b = 
 
3. 4
6
2 3 4 5
−
a b ab b = 
4. 2 23 2 2 3⋅ ⋅ ⋅ ⋅x x xa a a = 
 
5. ( )x y z w z zw z⋅ ⋅ ⋅2 3 0 2 3 4 2 = 6. 2 4 4 2 40 2 2 2 1⋅ ⋅ ⋅ ⋅ = 
 
 
Mastering Algebra - Advanced Level 1.1b Operations with Positive Integer Exponents 
Hamilton Education Guides 12 
Case II Dividing Positive Integer Exponents 
Positive integer exponents are divided by one another using the exponent laws I through VI 
shown in Table 1.1-1. 
Case II Dividing Positive Integer Exponents 
Positive integer exponents are divided by one another using the following steps: 
Step 1 a. Apply the Division and/or the Negative Power Laws (Laws V and VI) from Table 
 1.1-1. 
 b. Group the exponential terms with similar bases. 
Step 2 Apply the Multiplication Law (Law I) from Table 1.1-1 and simplify the exponential 
expressions by adding the exponents with similar bases. 
Examples with Steps 
The following examples show the steps as to how positive integer exponents are divided by one 
another: 
Example 1.1-22 
 2
4 3 4
a b
a b−
 = 
Solution: 
 Step 1 2
4 3 4
a b
a b−
 = − 2
4
1 1
3 4
a b
a b
 = ( ) ( )
−
⋅− −
2
4
1
3 1 4 1a a b b
 
 
 Step 2 ( ) ( )
−
⋅− −
2
4
1
3 1 4 1a a b b
 = ( ) ( )
−
/
/ ⋅− −
2
1
4
2
1
3 1 4 1a b
 = −
⋅
1
2
1
2 3a b
 = − 


1
2
1
2 3a b
 
Example 1.1-23 
 −
−
3
15
4 3
2 2
x y z y
x z
 = 
Solution: 
 Step 1 −
−
3
15
4 3
2 2
x y z y
x z
 = −
−
3
15
1 4 3 1
2 2
x y z y
x z
 = + 3
15
1 4 3 1
2 2
x y z y
x z
 = 
( ) ( )3
15
4 1 3 2
2 1
y y z z
x x
⋅ −
− 
 
 Step 2 
( ) ( )3
15
4 1 3 2
2 1
y y z z
x x
⋅ −
− = 
( ) ( )/
/ /
⋅+ −
−
3
1
15
5
4 1 3 2
2 1
y z
x
 = 1
5
5 1
1
y z
x
⋅ = 15
5y z
x





 
 
Additional Examples - Dividing Positive Integer Exponents 
The following examples further illustrate how to divide positive integer exponential terms by one 
another: 
Mastering Algebra - Advanced Level 1.1b Operations with Positive Integer Exponents 
Hamilton Education Guides 13 
Example 1.1-24 
x y z
x y
4 3
2 2 = 
( ) ( )x x y y z4 2 3 2
1
− −⋅ ⋅
 = 
( ) ( )x y z4 2 3 2
1
− −⋅ ⋅
 = x y z
2 1
1
⋅ ⋅ = x y z
2
1
 = x y z2 
Example 1.1-25 
5
15
3 5
2 0
a b
a b
 = 5
15 1
3 5
2
a b
a ⋅
 = 
( )/
/ /
⋅−5
1
15
3
1
3 2 5a a b
 = 
( )1
3 1
3 2 5a b− ⋅
 = 1
3 1
1 5a b⋅ = 13
5a b 
Example 1.1-26 
8
2
3 3 2
3 2
u w z
u w z
 = 8
2
3 3 2
3 2 1
u w z
u w z
 = 
( ) ( )
( )
/⋅ ⋅ ⋅
/⋅
− −
−
8
4
2
1
3 2 2 1
3 3
w w z z
u u
 = 
( ) ( )
( )
4
1
3 2 2 1
3 3
⋅ ⋅
⋅
− −
−
w z
u
 = 4
1 1
0
⋅ ⋅w z
u
 = 4
1
w z = 4w z 
Example 1.1-27 
100
5
2 2
4 5
p t u
p t u
 = 100
5
2 2 1
1 4 5
p t u
p t u
 = ( ) ( )
/ / /
/ ⋅
−
− −
100
20
5
1
2 1
4 2 5 1
p p
t t u u
 = ( ) ( )
20
1
2 1
4 2 5 1
p
t u
−
− −⋅
 = 20
1
1
2 4
p
t u⋅





 = 
20
2 4
p
t u
 
Example 1.1-28 
w z
z
w
z
2 2
3





 ⋅





 = w z
z
w
z
2 2
1
1
3





 ⋅





 = 
w z w
z z
2 2 1
1 3
⋅
⋅
 = w w
z z z
2 1
1 3 2− = 
w
z
2 1
1 3 2
+
+ − = 
w
z
3
2 
Practice Problems - Dividing Positive Integer Exponents 
Section 1.1b Case II Practice Problems - Divide the following positive integer exponents: 
 
1. x
x
5
3 = 
 
2. a b
a
2 3
 = 3. a b c
a b c
3 3 2
2 6 = 
4. 
( )
( )
3
2
2 2
3
⋅
⋅
r s
r s r
 = 
 
5. 2
6
2 3 4
4 2
p q p r
p q r−
 = 6. 
( ) ( )k l k l m
k l m
2 3 2 0
4 3 5
⋅
 = 
 
 
Mastering Algebra - Advanced Level 1.1b Operations with Positive Integer Exponents 
Hamilton Education Guides 14 
Case III Adding and Subtracting Positive Integer Exponents 
A common mistake among students is dealing with addition and subtraction of exponential 
expressions. In this section positive integer exponents addressing addition and subtraction of 
numbers that are raised to positive exponents is introduced. Positive exponential expressions are 
added and subtracted using the following steps: 
Step 1 Group the exponential terms with similar bases. 
Step 2 Simplify the exponential expressions by adding or subtracting the like terms. 
Note that like terms are defined as terms having the same variables raised to the same power. 
For example, x and x3 32 ; y and y2 24 are like terms of one another. 
Examples with Steps 
The following examples show the steps as to how exponential expressions having positive 
integer exponents are added or subtracted: 
Example 1.1-29 
 x y x y3 2 3 23 2 5+ + − + = 
Solution: 
 Step 1 x y x y3 2 3 23 2 5+ + − + = ( ) ( )x x y y3 3 2 22 3 5+ + − + 
 
 Step 2 ( ) ( )x x y y3 3 2 22 3 5+ + − + = ( ) ( )1 2 3 1 53 2+ + − +x y = 3 2 53 2x y+ + 
Example 1.1-30 
 ( ) ( )2 4 3 23 2 2 2+ + − + +x y x y x = 
Solution: 
 Step 1 ( ) ( )2 4 3 23 2 2 2+ + − + +x y x y x = 8 4 3 22 2 2+ + − − +x y x y x 
 
 = ( ) ( )x x x y y2 2 22 3 4 8+ − + − + 
 
 Step 2 ( ) ( )x x x y y2 2 22 3 4 8+ − + − + = ( ) ( )1 2 3 4 1 82+ − + − +x y = 0 3 82x y+ + = 3 8y + 
 
Additional Examples - Adding and Subtracting Positive Integer Exponents 
The following examples further illustrate addition and subtraction of exponential terms: 
Example 1.1-31 
Mastering Algebra - Advanced Level 1.1b Operations with Positive Integer Exponents 
Hamilton Education Guides 15 
5 3 2 53 2 3 2x x x x+ + − + = 5 3 2 3 3 2 2 5x x x x+

 + −



 + = ( ) ( )5 2 3 1 5
3 2+ + − +x x = 7 2 53 2x x+ + 
Example 1.1-32 
( ) ( )− − + + − − + +2 3 2 3 10 5 2 34 2 4 2m m m m m m = − − + + − − − −2 3 2 3 10 5 2 34 2 4 2m m m m m m 
 
= ( ) ( ) ( ) ( )− + + − − + − + − −2 2 3 5 3 2 10 34 4 2 2m m m m m m = ( ) ( ) ( )− + + − − + − −2 2 3 5 3 2 134 2m m m 
 
= 0 8 134 2m m m− + − = 8 132m m+ − 
Example 1.1-33 
x x y x y x x2 2 2 2 2 23 4 5 2 6+ + + − − + + = ( ) ( ) ( )x x x y y x x2 2 2 2 23 2 4 6 25+ + + − + + − 
 
= ( ) ( ) ( )1 3 2 1 4 1 6 252 2+ + + − + + −x y x = 6 3 7 252 2x y x− + − 
Example 1.1-34 
( ) ( )− − − − − − + +5 3 5 3 4 5 23 3w w w w w = ( ) ( )− − − + − + + + +5 3 5 3 4 5 23 3w w w w w 
 
= ( ) ( ) ( )− − + − + + + − + +5 3 3 5 5 4 23 3w w w w w = ( ) ( )− − + − + + +5 3 3 5 1 13w w = − + +8 3 13w w 
Example 1.1-35 
( ) ( )a b a b a b a2 3 2 4 2 3 2 0 33 2 2 2 3 3+ − + + + + − = a b a b a b a2 3 2 2 3 23 16 2 2 1 27+ − + + + + − 
 
= ( ) ( ) ( )a b a b a a b2 3 2 3 2 22 3 2 16 1 27+ + + − + + − = ( ) ( )1 2 3 2 102 3 2+ + + − −a b a b = 3 5 102 3 2a b a b+ − − 
 
Practice Problems - Adding and Subtracting Positive Integer Exponents 
Section 1.1b Case III Practice Problems - Add or subtract the following positive integer 
exponential expressions: 
1. x xy x xy z2 2 34 2 2+ − − + = 
 
2. ( ) ( )a a a3 2 3 32 4 4 20+ + − + = 
 
3. ( )3 2 2 2 34 2 4 4 2x x x x x+ + − − + = 
 
4. ( ) ( )− − + − − −2 2 5 4 203 3 2 2 3 3 3l a l a l a = 
 
5. ( ) ( )m m m m mn n n n3 2 3 24 2 3 5− − + + = 
 
6. ( ) ( )− + − − − + − + +7 3 5 3 4 5 203 3z z z z z = 
 
 
 
Mastering Algebra - Advanced Level 1.1c Operations with Negative Integer Exponents 
Hamilton Education Guides 16 
1.1c Operations with Negative Integer Exponents 
To proceed with simplification of negative exponents, we need to know the Negative Power Law 
in addition to the other exponent laws (shown in Table 1.1-2). The Negative Power Law states 
that a base raised to a negative exponent is equal to one divided by the same base raised to the 
positive exponent, or vice versa, i.e., 
a
a
n
n
− =
1 
and 
a
a
n
n= −
1 since a
a
a a
a a an n
n n
n n
n= = = =
×
×
= =−
1 1
1
1
1
1
1
1 1 1
. 
Note that the objective is to write the finalanswer without a negative exponent. To achieve this 
the exponent laws are used when simplifying expressions that have negative integer exponents. 
These laws are used to simplify the work in solving exponential expressions and should be 
memorized. 
Table 1.1-2: Exponent Laws 1 through 6 (Negative Integer Exponents) 
I. Multiplication a a am n m n− − − −⋅ = When multiplying negative exponential terms, if 
 bases a are the same, add the negative exponents 
 −m and −n . 
II. Power of a Power ( )a am n m n− − − ×−= When raising a negative exponential term to a 
 negative power, multiply the negative powers 
 (exponents) −m and −n . 
III. Power of a Product ( )a b a bm m m⋅ = ⋅− − − When raising a product to a negative power, raise 
 each factor a and b to the negative power −m . 
IV. Power of a Fraction a
b
a
b
m m
m




=
− −
−
 When raising a fraction to a negative power, raise 
 the numerator and the denominator to the negative 
 power −m . 
V. Division a
a
a a a
m
n
m n m n
−
−
− − += ⋅ = When dividing negative exponential terms, if the 
 bases a are the same, add exponents −m and n . 
VI. Negative Power a
a
n
n
− =
1 A non-zero base a raised to the −n power equals 1 
 divided by the base a to the n power 
 
In this section students learn how to multiply (Case I), divide (Case II), and add or subtract (Case 
III) negative integer exponents by one another. 
Case I Multiplying Negative Integer Exponents 
Negative integer exponents are multiplied by one another using the following steps and the 
exponent laws I through III shown in Table 1.1-2. 
Step 1 Group the exponential terms with similar bases. 
Mastering Algebra - Advanced Level 1.1c Operations with Negative Integer Exponents 
Hamilton Education Guides 17 
Step 2 Apply the Multiplication Law (Law I) from Table 1.1-2 and simplify the exponential 
expressions by adding the exponents with similar bases. 
Step 3 Change the negative integer exponents to positive integer exponents. 
Examples with Steps 
The following examples show the steps as to how negative integer exponents are multiplied by 
one another: 
Example 1.1-36 
 3 2 3 22 1 3− − −⋅ ⋅ ⋅ = 
Solution: 
 Step 1 3 2 3 22 1 3− − −⋅ ⋅ ⋅ = 3 2 3 22 1 1 3− − −⋅ ⋅ ⋅ = ( ) ( )3 3 2 22 1 3 1− − −⋅ ⋅ ⋅ 
 
 Step 2 ( ) ( )3 3 2 22 1 3 1− − −⋅ ⋅ ⋅ = ( ) ( )3 22 1 3 1− − − +⋅ = 3 23 2− −⋅ 
 
 Step 3 3 23 2− −⋅ = 1
3 23 2⋅
 = ( ) ( )
1
3 3 3 2 2⋅ ⋅ ⋅ ⋅
 = 1108 
Example 1.1-37 
 5 52 3 3 1− − − −⋅ ⋅ ⋅ ⋅ ⋅a b a b = 
Solution: 
 Step 1 5 52 3 3 1− − − −⋅ ⋅ ⋅ ⋅ ⋅a b a b = 5 52 1 3 3 1 1− − − −⋅ ⋅ ⋅ ⋅ ⋅a b a b = ( ) ( ) ( )5 52 1 3 1 1 3− − − −⋅ ⋅a a b b 
 
 Step 2 ( ) ( ) ( )5 52 1 3 1 1 3− − − −⋅ ⋅a a b b = ( ) ( ) ( )5 2 1 3 1 1 3− + − − −⋅ ⋅a b = 5 1 4 2− − −⋅ ⋅a b = 5 1 4 2− − −a b 
 
 Step 3 5 1 4 2− − −a b = 1
5
1 1
1 4 2⋅ ⋅a b
 = 1 1 1
5 4 2
⋅ ⋅
⋅ ⋅a b
 = 1
5 4 2a b
 
Additional Examples - Multiplying Negative Integer Exponents 
The following examples further illustrate how to multiply negative exponential terms by one 
another: 
Example 1.1-38 
a a a− −⋅ ⋅2 3 1 = a− + −2 3 1 = a0 = 1 
Example 1.1-39 
3 3 2 3 22 1 2 4− − − −⋅ ⋅ ⋅ ⋅ = 3 3 2 3 22 1 2 4 1− − − −⋅ ⋅ ⋅ ⋅ = ( ) ( )3 3 3 2 22 1 4 1 2− − − −⋅ ⋅ ⋅ ⋅ = ( ) ( )3 22 1 4 1 2− − − −⋅ = 3 27 1− −⋅ 
Mastering Algebra - Advanced Level 1.1c Operations with Negative Integer Exponents 
Hamilton Education Guides 18 
= 1
3
1
27 1
⋅ = 1
3 3 3 3 3 3 3
1
2⋅ ⋅ ⋅ ⋅ ⋅ ⋅
⋅ = 1
2187
1
2
⋅ = 1 1
2187 2
⋅
⋅
 = 14374 
Example 1.1-40 
a b a b− − − −⋅ ⋅ ⋅5 3 1 2 = ( ) ( )a a b b− − − −⋅ ⋅ ⋅5 1 3 2 = ( ) ( )a b− − − −⋅5 1 3 2 = a b− −⋅6 5 = a b− −6 5 = 16 5a b 
Example 1.1-41 
( ) ( )5 5 2 53 1 3 2− − − −⋅ ⋅ ⋅ = 5 5 2 53 1 3 2− − − −⋅ ⋅ ⋅ = ( )5 5 5 23 1 2 3− − − −⋅ ⋅ ⋅ = ( )5 23 1 2 3− − − −⋅ = 5 26 3− −⋅ 
 
= 1
5
1
26 3
⋅ = ( ) ( )
1
5 5 5 5 5 5
1
2 2 2⋅ ⋅ ⋅ ⋅ ⋅
⋅
⋅ ⋅
 = 1
15625
1
8
⋅ = ( )
1 1
15625 8
⋅
⋅
 = 1125000 
Example 1.1-42 
( ) ( ) ( )− + ⋅− − − −1 3 2 2 2 3 2 3r s t r s t s = ( )2 2 2 2 1 3 2 3 1− − − −⋅r s t r s t s = ( ) ( ) ( )2 2 2 3 2 2 1 1 3− − − −⋅ ⋅r r s s s t t 
 
= ( ) ( ) ( )2 2 2 3 2 2 1 1 3− − + − + −⋅ ⋅r s t = 2 2 1 1 2− −⋅ ⋅r s t = 2 2 2− −⋅ ⋅ ⋅r s t = 1
2 1 1
1
2 2⋅ ⋅ ⋅
r s
t
 = 1 1
2 2 1 1 2
⋅ ⋅ ⋅
⋅ ⋅ ⋅ ⋅
r s
t
 = r s
t4 2
 
Practice Problems - Multiplying Negative Integer Exponents 
Section 1.1c Case I Practice Problems - Multiply the following exponential expressions by one 
another: 
1. ( ) ( )3 2 2 3 23 1 3 2− − − −⋅ ⋅ ⋅ ⋅ = 
 
2. a b a b a− − − −⋅ ⋅ ⋅ ⋅6 4 1 2 0 = 
 
3. ( ) ( )a b a b− − −⋅ ⋅ ⋅2 3 2 2 = 
4. ( ) ( ) ( )− ⋅− − − −2 4 2 2 3 2 1r s t r s t s = 
 
5. 4
5
2 2
4
2 5 4 3 2



−
− − −v v v = 6. 2 3 3 2 21 2 5 2 0− −⋅ ⋅ ⋅ ⋅ = 
 
 
Mastering Algebra - Advanced Level 1.1c Operations with Negative Integer Exponents 
Hamilton Education Guides 19 
Case II Dividing Negative Integer Exponents 
Negative integer exponents are divided by one another using the exponent laws I through VI 
shown in Table 1.1-2. These laws are used in order to simplify division of negative fractional 
exponents by each other. Negative integer exponents are divided by one another using the 
following steps: 
Step 1 a. Apply the Division and/or the Negative Power Laws (Laws V and VI) from Table 
1.1-2. 
 b. Group the exponential terms with similar bases. 
Step 2 Apply the Multiplication Law (Law I) from Table 1.1-2 and simplify the exponential 
expressions by adding the exponents with similar bases. 
Examples with Steps 
The following examples show the steps as to how negative integer exponents are divided by one 
another: 
Example 1.1-43 
 5
5
2
3
−
− = 
Solution: 
 Step 1 5
5
2
3
−
− = 
5 5
1
3 2⋅ − = 
 
 Step 2 5 5
1
3 2⋅ − = 5
1
3 2−
 = 5
1
1
 = 5 
Example 1.1-44 
 a b
a b
− −
− −
2 3
1 2 = 
Solution: 
 Step 1 a b
a b
− −
− −
2 3
1 2 = ( ) ( )
1
2 1 3 2a a b b− −⋅
 
 
 Step 2 ( ) ( )
1
2 1 3 2a a b b− −⋅
 = ( ) ( )
1
2 1 3 2a b− −⋅
 = 11 1a b⋅
 = 1ab 
 
Additional Examples - Dividing Negative Integer Exponents 
The following examples further illustrate how to divide negative integer exponents by one another: 
Example 1.1-45 
Mastering Algebra - Advanced Level 1.1c Operations with Negative Integer Exponents 
Hamilton Education Guides 20 
a c
ac
− −
−
2 3
4 = −
− −a c
a c
2 3
1 4 = ( ) ( )
−
⋅
1
1 2 4 3a a c c
 = ( ) ( )
−
⋅+ +
1
1 2 4 3a c
 = − 13 7a c
 
Example 1.1-46 
( )
( )
−
−
−
−
3
3
3
3 = 
( )
( )
−
− −
−
3
3
3
3 = 
( )
( )
−
−
3
3
3
3 = −
⋅ ⋅
− ⋅ − ⋅ −
3 3 3
3 3 3
 = −
−
27
27
 = + / /
/ /
27
1
27
1
 = 1
1
 = 1 
Example 1.1-47 
( )
( )
−
− −
−
−
2
2
4
3 = 
( )
( )
−
−
−
−
−
2
2
4
3 = 
( )
( )
−
−
−
2
2
3
4 = −
− ⋅ − ⋅ −
− ⋅ − ⋅ − ⋅ −
2 2 2
2 2 2 2
 = − −8
16
 = + /
/ /
8
1
16
2
 = 12 
Example 1.1-48 
( )
− ⋅
−
− −
− −
2
2
3 1
2 3
a
a
 = 
( )
−
⋅
−
− −
− −
2
2
3 1
2 3
a
a
 = 
( ) ( )
−
− ⋅ ⋅ −2
2
2 3 1
3
a a
 = 
( ) ( )
−
− ⋅ − ⋅
⋅ ⋅
−2 2
2 2 2
3 1a
 = − ⋅4
8
2a = − /
/
4
1
8
2
2a = − 12
2a 
Example 1.1-49 
a
a
2
3 22− −⋅
 = 2
1
3 2 2⋅ ⋅a a = 8
1
2 2⋅ +a = 8
1
4a = 8 4a 
 
Practice Problems - Dividing Negative Integer Exponents 
Section 1.1c Case II Practice Problems - Divide the following negative integer exponents: 
 
1. x x
x x
−2
3 0 = 
 
2. −
−
−
− −
2
6
2 3
1 2
a b
a b
 = 3. ( )
( )
− −
⋅ −
−
−
3
3 3
4
3 = 
4. 
( )
−
−
−
− −
3
3
3 3
2 2 3
y y w
y w
 = 
 
5. a b a y
a y
− − −
−
2 2 5 2
3 = 6. 
( )x y z y x
x y
⋅ ⋅ ⋅ −
− −
0 2
4 1 = 
 
 
Mastering Algebra - Advanced Level 1.1c Operations with Negative Integer Exponents 
Hamilton Education Guides 21 
Case III Adding and Subtracting Negative Integer Exponents 
Negative exponential expressions are added and subtractedusing the following steps: 
Step 1 Group the exponential terms with similar bases. 
Step 2 Apply the Negative Power Law (Law VI) from Table 1.1-2, i.e., change a n− to 1
a n
. 
Step 3 Simplify the exponential expression by: 
 a. Using the fraction techniques learned in Section 1.1, and 
 b. Using appropriate exponent laws such as the Multiplication Law (Law I) from 
Table 1.1-2. 
Examples with Steps 
The following examples show the steps as to how exponential expressions having negative 
integer exponents are added or subtracted: 
Example 1.1-50 
 3 33 2− −+ = 
Solution: 
 Step 1 Not Applicable 
 
 Step 2 3 33 2− −+ = 1
3
1
33 2
+ = 1
27
1
9
+ 
 Step 3 1
27
1
9
+ = ( ) ( )1 9 1 27
27 9
⋅ + ⋅
⋅
 = 9 27
243
+ = / /
/ / /
36
4
243
27
 = 427 
Example 1.1-51 
 x x x x− − − − −+ + − +1 2 1 2 22 4 5 = 
Solution: 
 Step 1 x x x x− − − − −+ + − +1 2 1 2 22 4 5 = ( ) ( )x x x x− − − − −+ + − +1 1 2 2 22 4 5 
 
 = ( ) ( )1 2 1 4 51 2 2+ + − +− − −x x = 3 3 51 2 2x x− − −− + 
 
 Step 2 3 3 51 2 2x x− − −− + = 3 3 1
51 2 2x x
− + = 3 3 1
52 2x x
− + 
 
 Step 3 3 3 1
52 2x x
− + = 3 3 1
252x x
−




+ = 
( ) ( )3 3 1
25
2
2
⋅ − ⋅
⋅








+
x x
x x
 = 3 3 1
25
2
1 2
x x
x
−




 ++ 
 
Mastering Algebra - Advanced Level 1.1c Operations with Negative Integer Exponents 
Hamilton Education Guides 22 
 = 3 3 1
25
2
3
x x
x
−
+ = 
( )25 3 3 1
25
2 3
3
⋅ − + ⋅
⋅
x x x
x
 = 75 75
25
2 3
3
x x x
x
− + = x x x
x
3 2
3
75 75
25
+ − 
 
 = 
( )x x x
x
2
3
75 75
25
+ −
 = ( )
x x
x x
2
3 1
75 75
25
+ −
−
 = x x
x
2
3 1
75 75
25
+ −
− = 
x x
x
2
2
75 75
25
+ − 
Additional Examples - Adding and Subtracting Negative Integer Exponents 
The following examples further illustrate addition and subtraction of negative exponential terms: 
Example 1.1-52 
( )x y+ −5 = 
( )
1
5x y+
 Note: ( )x y⋅ −5 = 
( )
1
5x y⋅
 = 15 5x y
 
Example 1.1-53 
a b a b− − − −− + +1 1 1 12 3 = ( ) ( )a a b b− − − −+ + −1 1 1 12 3 = ( ) ( )1 2 3 11 1+ + −− −a b = 3 21 1a b− −+ = 3 2a b+ 
= ( ) ( )3 2⋅ + ⋅
⋅
b a
a b
 = 2 3a ba b
+ 
Example 1.1-54 
a b c b− − − −+ + +1 2 3 23 = ( )a b b c− − − −+ + +1 2 2 33 = ( )a b c− − −+ + +1 2 31 3 = a b c− − −+ +1 2 34 
 
= ( )a b c− − −+ +1 2 34 = 1 4 12 3a b c+




+ = 
( ) ( )1 4 12
2 3
⋅ + ⋅
⋅








+
b a
a b c
 = b a
a b c
2
2 3
4 1+




 + 
 
= 
( )[ ] ( )b a c ab
a b c
2 3 2
2 3
4 1+ ⋅ + ⋅
⋅
 = b c a c a b
a b c
2 3 3 2
2 3
4⋅ + ⋅ + = b c a c a b
a b c
2 3 3 2
2 3
4+ + 
Example 1.1-55 
5 3 2 53 2 3 2 2x x x x− − − −+ + − + − = 5 2 3 53 3 2 2 2x x x x− − − − −+

 + −



 + = ( ) ( )5 2 3 1 5
3 2 2+ + − +− − −x x 
 
= 7 2 53 2 2x x− −+ + − = 7 2 1
53 2 2x x
+ + = 7 2 1
253 2x x
+





 + = 7 2 1
25
2 3
3 2
⋅ + ⋅
⋅







 +
x x
x x
 = 7 2 1
25
2 3
3 2
x x
x
+
+
+ 
 
= ( )x x
x
2
5
7 2 1
25
+
+ = 7 2 1
255 2
+
⋅
+
−
x
x x
 = 7 2 1
255 2
+
+
−
x
x
 = 7 2 1
253
+
+
x
x
 = 
( )[ ] ( )7 2 25 1
25
3
3
+ ⋅ + ⋅
⋅
x x
x
 
 
Mastering Algebra - Advanced Level 1.1c Operations with Negative Integer Exponents 
Hamilton Education Guides 23 
= 175 50
25
3
3
+ +x x
x
 = x x
x
3
3
50 175
25
+ + 
Example 1.1-56 
x x y x y x− − − −+ + + − − +2 2 2 2 2 2 23 4 5 2 = ( ) ( ) ( )x x y y x x− − − −+ + − + + −2 2 2 2 2 23 4 2 25 
 
= ( ) ( ) ( )1 3 1 4 1 2 252 2 2+ + − + + −− −x y x = 4 3 3 252 2 2x y x− −− + − = 4 3 3 252 2
2
x y
x−





 + − 
 
= 4 3 3 25
1
2 2
2 2
2⋅ − ⋅
⋅





 +
−y x
x y
x = 4 3 3 25
1
2 2
2 2
2y x
x y
x−
+
− = 
( )[ ] ( )[ ]4 3 1 3 252 2 2 2 2
2 2
y x x y x
x y
− ⋅ + −
 
 
= 4 3 3 25
2 2 4 2 2 2
2 2
y x x y x y
x y
− + − 
Practice Problems - Adding and Subtracting Negative Integer Exponents 
Section 1.1c Case III Practice Problems - Simplify the following negative integer exponential 
expressions: 
1. x x x x− − − −+ + −1 2 1 22 3 6 = 2. ( ) ( )3 2 34 2 4 2a b a b− − − −− + − + = 
 
3. ( ) ( )xy y xy y− − − − −+ + − +1 2 1 2 34 3 2 = 
 
4. 4 51 3 3x y y− − −+ + = 
 
5. ( )m m m m m− − − −− − + +5 2 5 0 23 3 = 
 
6. ( ) ( )a a b a a b3 2 2 2 6 4 26 3− − − −+ − + = 
 
Mastering Algebra - Advanced Level 1.2a Introduction to Radicals 
Hamilton Education Guides 24 
1.2 Radicals 
In this section radical expressions and the steps as to how they are simplified are reviewed in 
section 1.2a, Cases I through III. In addition, operations involving radical expressions which 
include multiplication, division (rationalization), addition, and subtraction of radicals are 
addressed in section 1.2b, Cases I through VI. 
1.2a Introduction to Radicals 
A description of roots and radicals (Case I), classification of numbers (Case II), and 
simplification of radical expressions with real numbers as a radicand (Case III) are discussed 
below: 
Case I Roots and Radical Expressions 
In the general radical expression b ca = , the symbol is called a radical sign. The expression 
under the radical b is called the radicand, a is called the index, and the positive square root of 
the number c is called the principal square root. 
Exponents are a kind of shorthand for multiplication. For example, 5 5 25× = can be expressed in 
exponential form as 5 252 = . Radical signs are used to reverse this process. For example, to 
write the reverse of 5 252 = we take the square root of the terms on both sides of the equal sign, 
i.e., we write 25 5 52= = . Note that since 5 252 = and ( )− =5 252 , we use 25 to indicate the 
positive square root of 25 is equal to 5 and − 25 to indicate the negative square root of 25 is 
equal to −5 . Table 1.2-1 provides square roots, cube roots, fourth roots, and fifth roots of some 
common numbers used in solving radical expressions. This table should be used as a reference 
when simplifying radical terms. Students are not encouraged to memorize this table. Following 
are a few examples on simplifying radical expressions using Table 1.2-1: 
a. 64 8 82= = b. ( )− = − = − ⋅ = −2 25 2 5 2 5 102 
c. ( )5 32 5 2 5 2 105 55= = ⋅ = d. 125 25 5 5 5 5 52= ⋅ = ⋅ = 
e. 147 49 3 7 3 7 32 2= ⋅ = ⋅ = f. ( )2 32 2 16 2 2 4 2 2 4 2 8 22= ⋅ = ⋅ = ⋅ = 
g. 2048 1024 2 4 2 4 25 5 55 5= ⋅ = ⋅ = h. 375 125 3 5 3 5 33 3 33 3= ⋅ = ⋅ = 
i. ( )2 250 2 25 10 2 5 10 2 5 10 10 102= ⋅ = ⋅ = ⋅ = j. 324 81 4 3 4 3 44 4 44 4= ⋅ = ⋅ = 
k. 648 216 3 6 3 6 33 3 33 3= ⋅ = ⋅ = l. ( )− = − ⋅ = − ⋅ = − ⋅ = −324 81 4 9 2 9 2 182 2 2 2 
Practice Problems - Roots and Radical Expressions 
Section 1.2a Case I Practice Problems - Simplify the following radical expressions by using 
Table 1.2-1: 
1. 982 = 
 
2. 3 75 = 
 
3. 1253 = 
 
4. 31255 = 
 
5. 1624 = 
 
6. 1922 = 
 
 
Mastering Algebra - Advanced Level 1.2a Introduction to Radicals 
Hamilton Education Guides 25 
 
Table 1.2-1: Square roots, cube roots, fourth roots, and fifth roots 
 
Square Roots Cube Roots 
( ) ( )1 1 1 1 12
1
2
2
1
2= = = = Note a a: 2 = ( ) ( )1 1 1 1 13 33
1
3
3
1
3= = = = 
( ) ( )4 2 4 2 22
1
2
2
1
2= = = = ( ) ( )8 2 8 23 33
1
3
3
1
3 2= = = = 
( ) ( )9 3 9 3 32
1
2
2
1
2= = = = ( ) ( )27 3 27 3 33 33
1
3
3
1
3= = = = 
( ) ( )16 4 16 4 42
1
2
2
1
2= = = = ( ) ( )64 4 64 4 43 33
1
3
3
1
3= = = = 
( ) ( )25 5 25 5 52
1
2
2
1
2= = = = ( ) ( )125 5 125 5 53 33
1
3
3
1
3= = = = 
( ) ( )36 6 36 6 62
1
2
2
1
2= = = = ( ) ( )216 6 216 6 63 33
1
3
3
1
3= = = = 
( ) ( )49 7 49 7 72
1
2
2
1
2= = = = ( ) ( )343 7 343 7 73 33
1
3
3
1
3= = = = 
( ) ( )64 8 64 8 82
1
2
2
1
2= = = = ( ) ( )512 8 512 8 83 33
1
3
3
1
3= = = = 
( ) ( )81 9 81 9 92
1
2
2
1
2= = = = ( ) ( )729 9 729 9 93 33
1
3
3
1
3= = = = 
( ) ( )100 10 100 10 102
1
2
2
1
2== = = ( ) ( )1000 10 1000 10 103 33
1
3
3
1
3= = = = 
Fourth Roots Fifth Roots 
( ) ( )1 1 1 1 14 44
1
4
4
1
4= = = = ( ) ( )1 1 1 1 15 55
1
5
5
1
5= = = = 
( ) ( )16 2 16 2 24 44
1
4
4
1
4= = = = ( ) ( )32 2 32 2 25 55
1
5
5
1
5= = = = 
( ) ( )81 3 81 3 34 44
1
4
4
1
4= = = = ( ) ( )243 3 243 3 35 55
1
5
5
1
5= = = = 
( ) ( )256 4 256 4 44 44
1
4
4
1
4= = = = ( ) ( )1024 4 1024 4 45 55
1
5
5
1
5= = = = 
( ) ( )625 5 625 5 54 44
1
4
4
1
4= = = = ( ) ( )3125 5 3125 5 55 55
1
5
5
1
5= = = = 
( ) ( )1296 6 1296 6 64 44
1
4
4
1
4= = = = ( ) ( )7776 6 7776 6 65 55
1
5
5
1
5= = = = 
( ) ( )2401 7 2401 7 74 44
1
4
4
1
4= = = = ( ) ( )16807 7 16807 7 75 55
1
5
5
1
5= = = = 
( ) ( )4096 8 4096 8 84 44
1
4
4
1
4= = = = ( ) ( )32768 8 32768 8 85 55
1
5
5
1
5= = = = 
( ) ( )6561 9 6561 9 94 44
1
4
4
1
4= = = = ( ) ( )59049 9 59049 9 95 55
1
5
5
1
5= = = = 
( ) ( )10000 10 10000 10 104 44
1
4
4
1
4= = = = ( ) ( )100000 10 100000 10 105 55
1
5
5
1
5= = = = 
 
Mastering Algebra - Advanced Level 1.2a Introduction to Radicals 
Hamilton Education Guides 26 
Case II Rational, Irrational, Real, and Imaginary Numbers 
A rational number is a number that can be expressed as: 
1. An integer fraction a
b
, where a and b are integer numbers and b ≠ 0 . For example: 
3
8
, − 4
5
, 25
100
, and 2
7
 are rational numbers. 
2. The square root of a perfect square, the cube root of a perfect cube, etc. For example: 
36 6 62= = , 49 7 72= = , − = − = −125 5 53 33 , 81 3 34 44= = , and − = − = −1024 4 45 55 are 
rational numbers. 
3. An integer (a whole number). For example: 5 5
1
= , 0 , 0
28
0= , and 10 are rational numbers. 
4. A terminating decimal. For example: 0 25 25
100
. = , −0 75. , and 5 5 5 1
2
. = − are rational numbers. 
5. A repeating decimal. For example: 0 3333333 1
3
. ...= , 0 45454545. ... , and are rational numbers. 
An irrational number is a number that: 
1. Can not be expressed as an integer fraction a
b
, where a and b are integer numbers and 
b ≠ 0 . For example: π , 2
2
, and − 5
3
 are irrational numbers. 
2. Can not be expressed as the square root of a perfect square, the cube root of a perfect cube, 
etc. For example: 5 , − 7 , 12 , 43 , − 65 , and 3 are irrational numbers. 
3. Is not a terminating or repeating decimal. For example: 0 432643. ..., −8 346723. ... , and 
314159. ... are irrational numbers. 
The real numbers consist of all the rational and irrational numbers. For example: π , 2
2
, 
− 7 , 43 , − 65 , 3 , 36 6 62= = , 0 25 25
100
. = , −0 75. , − = −55 5 1
2
. , − = −38 3 4
5
. , 5 5
1
= , 0 , and 
− = −25 25
1
 are real numbers. 
The not real numbers or imaginary numbers are square root of any negative real number. For 
example: −15 , −9 , −45 , and −36 are imaginary numbers. Note that imaginary numbers 
are also shown as i1515 =− , ii 399 ==− , i4545 =− , and ii 63636 ==− in advanced 
math books (See Chapter 2, Sections 2.5 and 2.6). 
Practice Problems - Rational, Irrational, Real, and Imaginary Numbers 
Section 1.2a Case II Practice Problems - Identify which one of the following numbers are 
rational, irrational, real, or not real: 
1. 5
8
 = 2. 45 = 
 
3. 450 = 
 
4. − 2
10
 = 5. − −5 = 
 
6. 5
2−
 = 
 
Mastering Algebra - Advanced Level 1.2a Introduction to Radicals 
Hamilton Education Guides 27 
Case III Simplifying Radical Expressions with Real Numbers as a Radicand 
Radical expressions with a real number as radicand are simplified using the following general 
rule: 
 a a ann
n
n= = The nth root of a n is a 
Where a is a positive real number and n is an integer. 
Radicals of the form a ann = are simplified using the following steps: 
Step 1 Factor out the radicand a n to a perfect square, cube, fourth, fifth, etc. term (use Table 
1.2-1). Write any term under the radical that exceeds the index n as multiple sum of 
the index. 
Step 2 Use the Multiplication Law for exponents (see Section 1.1) by writing a m n+ in the 
form of a am n⋅ . 
Step 3 Simplify the radical expression by using the general rule a ann = . Note that any term 
under the radical which is less than the index n stays inside the radical. 
Examples with Steps 
The following examples show the steps as to how radical expressions with real terms are 
simplified: 
Example 1.2-1 
 1
8
722
−
 = 
Solution: 
 Step 1 1
8
722
−
 = − 1
8
72 = − ⋅1
8
36 2 = ( )− ⋅ ⋅1
8
6 6 2 = ( )− ⋅ ⋅18 6 6 21 1 
 
 = ( )− ⋅+18 6 21 1 = − ⋅
1
8
6 22 
 
 Step 2 Not Applicable 
 
 Step 3 − ⋅1
8
6 22 = − ⋅1
8
6 2 = − /
/
6
3
8
2
4
 = − 34 2 
Example 1.2-2 
 − ⋅3
10
5 474 = 
Solution: 
 Step 1 − ⋅3
10
5 474 = − ⋅+3
10
5 44 34 
Mastering Algebra - Advanced Level 1.2a Introduction to Radicals 
Hamilton Education Guides 28 
 Step 2 − ⋅+3
10
5 44 34 = − ⋅ ⋅3
10
5 5 44 34 
 
 Step 3 − ⋅ ⋅3
10
5 5 44 34 = − ⋅ ⋅3
10
5 5 434 = − / /
/ /
⋅
15
3
10
2
5 434 = − ⋅3
2
125 44 = −
3
2
5004 
Additional Examples - Simplifying Radical Expressions with Real Numbers as a Radicand 
The following examples further illustrate how to solve radical expressions with real numbers as 
radicand: 
Example 1.2-3 
252 = 25 = 5 5⋅ = 5 51 1⋅ = 51 1+ = 52 = 5 
Example 1.2-4 
182 = 18 = 9 2⋅ = ( )3 3 2⋅ ⋅ = ( )3 3 21 1⋅ ⋅ = ( )3 21 1+ ⋅ = 3 22 ⋅ = 3 2 
Example 1.2-5 
1252 = 125 = 25 5⋅ = ( )5 5 5⋅ ⋅ = ( )5 5 51 1⋅ ⋅ = ( )5 51 1+ ⋅ = 5 52 ⋅ = 5 5 
Example 1.2-6 
24002 = 2400 = 400 6⋅ = ( )20 20 6⋅ ⋅ = ( )20 20 61 1⋅ ⋅ = ( )20 61 1+ ⋅ = 20 62 ⋅ = 20 6 
Example 1.2-7 
1202 = 120 = 4 30⋅ = ( )2 2 30⋅ ⋅ = ( )2 2 301 1⋅ ⋅ = ( )2 301 1+ ⋅ = 2 302 ⋅ = 2 30 
Example 1.2-8 
2
15
502 = 2
15
50 = 2
15
25 2⋅ = ( )2
15
5 5 2⋅ ⋅ = ( )215 5 5 21 1⋅ ⋅ = ( )
2
15
5 21 1+ ⋅ = 2
15
5 22 ⋅ 
 
= 2
15
5 2⋅ = / /
/ /
10
2
15
3
2 = 23 2 
 
Practice Problems - Simplifying Radical Expressions with Real Numbers as a Radicand 
Section 1.2a Case III Practice Problems - Simplify the following radical expressions: 
1. − 49 = 
 
2. 54 = 3. − 500 = 
4. 3 555 ⋅ = 5. 2162 = 6. − ⋅1
4
4 254 = 
 
Mastering Algebra - Advanced Level 1.2b Operations Involving Radical Expressions 
Hamilton Education Guides 29 
1.2b Operations Involving Radical Expressions 
In this section multiplication, division, addition, and subtraction of monomial and binomial 
radical expressions, with real numbers, is addressed in Cases I through VI. 
Case I Multiplying Monomial Expressions in Radical Form, with Real Numbers 
Radicals are multiplied by each other by using the following general product rule: 
( )a x b y c z a b c x y z abc xyzn n n n n⋅ ⋅ = ⋅ ⋅ ⋅ ⋅ = 
Note that radicals can only be multiplied by each other if they have the same index n . 
 A monomial expression in radical form is defined as: 
8 5x , y , 27 , −3 6 75 x y , x y2 53 , 2 125 , etc. 
 A binomial expression in radical form is defined as: 
x y+ , 1 8+ x , xy y+ 3 , x y x y3 3 2− , 9 5 65− x y , 643 2 53+ x y , etc. 
Monomial expressions in radical form are multiplied by each other using the above general 
product rule. Radical expressions with real numbers as radicands are multiplied by each other 
using the following steps: 
Step 1 Simplify the radical terms (see Section 1.2a, Case III). 
Step 2 Multiply the radical terms by using the product rule. Repeat Step 1, if necessary. 
( )k a k b k c k k k a b c k k k abcn n n n n1 2 3 1 2 3 1 2 3⋅ ⋅ = ⋅ ⋅ ⋅ ⋅ = a b and c, , ≥ 0 
Examples with Steps 
The following examples show the steps as to how radical expressions in monomial form are 
multiplied by one another: 
Example 1.2-9 
 5 15⋅ = 
Solution: 
 Step 1 Not Applicable 
 
 Step 2 5 15⋅ = 5 15⋅ = 75 = 25 3⋅ = 5 32 ⋅ = 5 3 
Example 1.2-10 
 98 48 108⋅ ⋅ = 
Solution: 
 Step 1 98 48 108⋅ ⋅ = 49 2 16 3 36 3⋅ ⋅ ⋅ ⋅ ⋅ = 7 2 4 3 6 32 2 2⋅ ⋅⋅ ⋅ ⋅ 
 
 = 7 2 4 3 6 3⋅ ⋅ 
 
 Step 2 7 2 4 3 6 3⋅ ⋅ = ( ) ( )7 4 6 2 3 3⋅ ⋅ ⋅ ⋅ ⋅ = ( )168 2 3 3⋅ ⋅ = 168 2 32⋅ 
Mastering Algebra - Advanced Level 1.2b Operations Involving Radical Expressions 
Hamilton Education Guides 30 
 = ( )168 3 2⋅ = 504 2 
 
Additional Examples - Multiplying Monomial Expressions in Radical Form, with Real Numbers 
The following examples further illustrate how to multiply radical terms by one another: 
Example 1.2-11 
12 9⋅ = ( )4 3 32⋅ ⋅ = 2 3 32 ⋅  ⋅ = 2 3 3⋅ = ( )2 3 3⋅ = 6 3 
Example 1.2-12 
20 50⋅ = 4 5 25 2⋅ ⋅ ⋅ = 2 5 5 22 2⋅ ⋅ ⋅ = 2 5 5 2⋅ = ( ) ( )2 5 5 2⋅ ⋅ ⋅ = 10 10 
Note that we can also simplify the radical terms in the following way: 
20 50⋅ = 20 50⋅ = 1000 = 100 10⋅ = 10 102 ⋅ = 10 10 
Example 1.2-13 
50 32 3⋅ ⋅ = 25 2 16 2 3⋅ ⋅ ⋅ ⋅ = 5 2 4 2 32 2⋅ ⋅ ⋅ ⋅ = 5 2 4 2 3⋅ ⋅ = ( ) ( )5 4 2 2 3⋅ ⋅ ⋅ ⋅ 
 
= ( )20 2 2 3⋅ ⋅ ⋅ = 20 2 2 31 1⋅ ⋅ ⋅  = 20 2 31 1⋅ ⋅ + = 20 2 32 ⋅ = ( )20 2 3⋅ = 40 3 
Example 1.2-14 
128 5003 3⋅ = 2 64 4 1253 3⋅ ⋅ ⋅ = 2 4 4 533 33⋅ ⋅ ⋅ = 4 2 5 43 3⋅ = ( ) ( )4 5 2 43 3⋅ ⋅ ⋅ = ( )20 2 43⋅ ⋅ 
 
= 20 83⋅ = 20 233⋅ = 20 2⋅ = 40 
Example 1.2-15 
54 484 4⋅ = 2 27 3 164 4⋅ ⋅ ⋅ = 2 3 3 234 44⋅ ⋅ ⋅ = 2 3 2 334 14⋅ ⋅ = 2 2 3 33 14 ⋅ ⋅ = 2 2 33 14 ⋅ + = 2 2 344 ⋅ 
 
= ( )2 3 24⋅ ⋅ = 6 24 
Practice Problems - Multiplying Monomial Expressions in Radical Form, with Real Numbers 
 
Section 1.2b Case I Practice Problems - Multiply the following radical expressions: 
 
1. 72 75⋅ = 
 
2. − ⋅3 20 2 32 = 3. 16 272 2⋅ = 
4. 64 100 54⋅ ⋅ = 
 
5. − ⋅ −125 2 98 = 6. 625 324 484 4 4⋅ ⋅ = 
 
Mastering Algebra - Advanced Level 1.2b Operations Involving Radical Expressions 
Hamilton Education Guides 31 
Case II Multiplying Binomial Expressions in Radical Form, with Real Numbers 
To multiply two binomial radical expressions the following multiplication method known as the 
FOIL method needs to be memorized: 
( ) ( ) ( ) ( ) ( ) ( )a b c d a c a d b c b d+ ⋅ + = ⋅ + ⋅ + ⋅ + ⋅ 
Multiply the First two terms, i.e., ( )a c⋅ . 
Multiply the Outer two terms, i.e., ( )a d⋅ . 
Multiply the Inner two terms, i.e., ( )b c⋅ . 
Multiply the Last two terms, i.e., ( )b d⋅ . 
Examples: 
 
1. ( ) ( )u v u v+ ⋅ − = ( ) ( ) ( ) ( )u u u v v u v v⋅ − + ⋅ − ⋅ = ( ) ( ) ( ) ( )u u u v u v v v⋅ − ⋅ + ⋅ − ⋅ 
 
 = u v2 2− = u v− 
 
2. ( ) ( )3 5 5 7− ⋅ + = ( ) ( ) ( ) ( )3 5 3 7 5 5 5 7⋅ + ⋅ − ⋅ − ⋅ = 15 3 7 5 5 5 7+ − − ⋅ = 15 3 7 5 5 35+ − − 
Binomial radical expressions are multiplied by each other using the following steps: 
Step 1 Simplify the radical terms (see Section 1.2a, Case III). 
Step 2 Use the FOIL method to multiply each term. Repeat Step 1, if necessary. 
( ) ( ) ( ) ( ) ( ) ( )a b c d a c a d b c b d+ ⋅ + = ⋅ + ⋅ + ⋅ + ⋅ 
Examples with Steps 
The following examples show the steps as to how binomial radical expressions with real 
numbers as radicands are multiplied by one another: 
Example 1.2-16 
 ( ) ( )2 2 5 8+ ⋅ − = 
Solution: 
 Step 1 ( ) ( )2 2 5 8+ ⋅ − = ( ) ( )2 2 5 4 2+ ⋅ − ⋅ = ( )2 2 5 2 22+ ⋅ − ⋅

 
 
 = ( ) ( )2 2 5 2 2+ ⋅ − 
 
 Step 2 ( ) ( )2 2 5 2 2+ ⋅ − = ( ) ( ) ( ) ( )2 5 2 2 2 5 2 2 2 2⋅ − ⋅ + ⋅ − ⋅ 
 
 = 10 4 2 5 2 2 2 2− + − ⋅ = 10 4 2 5 2 2 22− + − = ( ) ( )10 4 5 2 2 2+ − + − ⋅ 
 
Mastering Algebra - Advanced Level 1.2b Operations Involving Radical Expressions 
Hamilton Education Guides 32 
 = 10 2 4+ − = ( )10 4 2− + = 6 2+ 
Example 1.2-17 
 ( ) ( )24 3 60 25 72+ ⋅ − = 
Solution: 
 Step 1 ( ) ( )24 3 60 25 72+ ⋅ − = ( ) ( )4 6 3 4 15 5 5 36 2⋅ + ⋅ ⋅ ⋅ − ⋅ 
 
 = 2 6 3 2 15 5 6 22 2 2 2⋅ + ⋅

 ⋅ − ⋅



 = ( )( ) ( )2 6 3 2 15 5 6 2+ ⋅ ⋅ − 
 
 = ( ) ( )2 6 6 15 5 6 2+ ⋅ − 
 
 Step 2 ( ) ( )2 6 6 15 5 6 2+ ⋅ − = ( ) ( ) ( ) ( )5 2 6 2 6 6 2 5 6 15 6 15 6 2⋅ − ⋅ + ⋅ − ⋅ 
 
 = ( ) ( )10 6 2 6 6 2 30 15 6 6 15 2− ⋅ ⋅ + − ⋅ ⋅ = 10 6 12 12 30 15 36 30− + − 
 
 = 10 6 12 4 3 30 15 36 30− ⋅ + − = 10 6 12 2 3 30 15 36 302− ⋅ + − 
 
 = ( )10 6 12 2 3 30 15 36 30− ⋅ + − = 10 6 24 3 30 15 36 30− + − 
 
Additional Examples - Multiplying Binomial Expressions in Radical Form, with Real Numbers 
The following examples further illustrate how to multiply radical expressions by one another: 
Example 1.2-18 
( ) ( )3 300 8 50+ ⋅ − = ( ) ( )3 100 3 8 25 2+ ⋅ ⋅ − ⋅ = 3 10 3 8 5 22 2+ ⋅  ⋅ − ⋅  = ( ) ( )3 10 3 8 5 2+ ⋅ − 
 
= ( ) ( ) ( ) ( )3 8 3 5 2 8 10 3 10 5 3 2⋅ − ⋅ + ⋅ − ⋅ ⋅ = 24 15 2 80 3 50 3 2− + − ⋅ = − + − +15 2 80 3 50 6 24 
Example 1.2-19 
( ) ( )3 12 75 2+ ⋅ − = ( ) ( )3 4 3 25 3 2+ ⋅ ⋅ ⋅ − = 3 2 3 5 3 22 2+ ⋅  ⋅ ⋅ −  
 
= ( ) ( )3 2 3 5 3 2+ ⋅ − = ( ) ( ) ( )( ) ( )3 5 3 3 2 2 5 3 3 2 3 2⋅ − ⋅ + ⋅ ⋅ − ⋅ = 15 3 3 2 10 3 3 2 3 2− + ⋅ − ⋅ 
 
= 15 3 3 2 10 3 2 62− + − = ( )15 3 3 2 10 3 2 6− + ⋅ − = 15 3 3 2 2 6 30− − + 
Example 1.2-20 
3 5 4 2 3 2 233 33 3 33⋅ − ⋅

 ⋅ − ⋅



 = ( ) ( )5 3 2 4 3 2 23 3 3 3− ⋅ − 
Mastering Algebra - Advanced Level 1.2b Operations Involving Radical Expressions 
Hamilton Education Guides 33 
= ( ) ( )( ) ( ) ( )( )5 3 3 5 2 3 2 2 4 3 2 2 4 23 3 3 3 3 3 3 3⋅ − ⋅ ⋅ − ⋅ + ⋅ ⋅ = ( ) ( ) ( ) ( )5 3 3 10 3 2 2 4 3 4 4 23 3 3 3⋅ − ⋅ − ⋅ + ⋅ 
 
= 5 9 10 6 2 12 4 83 3 3 3− − + = 5 9 10 6 2 12 4 23 3 3 33− − + = ( )5 9 10 6 2 12 4 23 3 3− − + ⋅ 
 
= 5 9 10 6 2 12 83 3 3− − + 
Example 1.2-21 
( ) ( )6 48 2 2 18 4+ ⋅ − = ( ) ( )6 3 16 2 2 2 9 4⋅ + ⋅ ⋅ − = 6 3 4 2 2 2 3 42 2⋅ +


⋅ ⋅ −


 
 
= ( ) ( )6 4 3 2 2 3 2 4⋅ + ⋅ ⋅ − = ( ) ( )24 3 2 6 2 4+ ⋅ − = ( ) ( ) ( ) ( )24 3 6 2 4 24 3 2 6 2 2 4⋅ − ⋅ + ⋅ − ⋅ 
 
= ( ) ( ) ( )24 6 3 2 96 3 12 2 8⋅ ⋅ − + − = 144 6 96 3 12 2 8− + − 
Example 1.2-22 
( ) ( )[ ] ( )− + ⋅ − ⋅ −3 2 3 3 3 4 = ( ) ( ) ( ) ( )[ ] ( )− ⋅ + ⋅ + ⋅ − ⋅ ⋅ −3 3 3 3 2 3 2 3 3 4 
 
= [ ] ( )− + ⋅ + − ⋅ −3 3 3 3 6 2 3 3 4 = ( ) ( )− − + +


⋅ −3 3 2 3 3 6 3 42 = [ ] ( )− + + ⋅ −5 3 3 6 3 4 
 
= [ ] ( )− + ⋅ −5 3 9 3 4 = ( ) ( ) ( )− ⋅ + ⋅ + ⋅ − ⋅5 3 3 5 4 3 9 3 9 4 = − ⋅ + + −5 3 3 20 3 9 3 36 
 
= ( )− + + −5 3 20 9 3 362 = ( )− ⋅ + −5 3 29 3 36 = − − +15 36 29 3 = − +51 29 3 
Practice Problems - Multiplying Binomial Expressions in Radical Form, with Real Numbers 
Section 1.2b Case II Practice Problems - Multiply the following radical expressions: 
 
1. ( ) ( )2 3 1 2 2+ ⋅ + = 
 
2. ( ) ( )1 5 8 5+ ⋅ + = 3. ( ) ( )2 2 3 2− ⋅ + = 
4. ( )5 5 5 53+ ⋅ −  = 
 
5. ( ) ( )2 6 16 184+ ⋅ − = 6. ( ) ( )2 5 45 814− ⋅ + = 
 
 
Mastering Algebra - Advanced Level 1.2b Operations Involving Radical Expressions 
Hamilton Education Guides 34 
Case III Multiplying Monomial and Binomial Expressions in Radical Form, with Real Numbers 
To multiply monomial and binomial expressions in radical form the following general 
multiplication rule is used: 
( )a b c a b a c⋅ + = ⋅ + ⋅ 
Monomial and binomial expressions in radical form are multiplied by each other using the 
following steps: 
Step 1 Simplify the radical terms (see Section 1.2a, Case III). 
Step 2 Multiply each term using the general multiplication rule, i.e., ( )a b c a b a c⋅ + = ⋅ + ⋅ . 
Repeat Step 1, if necessary. 
Examples with Steps 
The following examples show the steps as to how monomial and binomial expressions in radical 
form are multiplied by one another: 
Example 1.2-23 
 ( )5 50 2 27⋅ + = 
Solution: 
 Step 1 ( )5 50 2 27⋅ + = ( )5 25 2 2 9 3⋅ ⋅ + ⋅ = 5 5 2 2 3 32 2⋅ ⋅ + ⋅  
 
 = ( )[ ]5 5 2 2 3 3⋅ + ⋅ = [ ]5 5 2 6 3⋅ + 
 
 Step 2 [ ]5 5 2 6 3⋅ + = ( ) ( )5 2 5 6 5 3⋅ + ⋅ = 5 2 5 6 5 3⋅ + ⋅ = 5 10 6 15+ 
 
Example 1.2-24 
 ( )− ⋅ −2 4 64 1624 4 4 = 
Solution: 
 Step 1 ( )− ⋅ −2 4 64 1624 4 4 = ( )− ⋅ ⋅ − ⋅2 4 16 4 81 24 4 4 = − ⋅ ⋅ − ⋅ 2 4 2 4 3 24 44 44 
 
 = ( )− ⋅ −2 4 2 4 3 24 4 4 
 
 Step 2 ( )− ⋅ −2 4 2 4 3 24 4 4 = ( ) ( ) ( ) ( )− ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅2 2 4 4 2 3 4 24 4 4 4= ( ) ( )− ⋅ + ⋅4 4 4 6 4 24 4 
 
 = − +4 16 6 84 4 = − +4 2 6 844 4 = ( )− ⋅ +4 2 6 84 = − +8 6 84 = ( )2 3 8 44 − 
 
 
 
Mastering Algebra - Advanced Level 1.2b Operations Involving Radical Expressions 
Hamilton Education Guides 35 
Additional Examples - Multiplying Monomial and Binomial Expressions in Radical Form, with Real Numbers 
The following examples further illustrate how to multiply radical terms by one another: 
Example 1.2-25 
( )3 5 5 6 10⋅ + = ( ) ( )3 5 5 3 5 6 10⋅ + ⋅ = ( ) ( )3 5 5 3 6 5 10⋅ + ⋅ ⋅ = 3 5 18 502 + 
 
= ( )3 5 18 25 2⋅ + ⋅ = 15 18 5 22+ ⋅ = ( )15 18 5 2+ ⋅ = 15 90 2+ = ( )15 1 6 2+ 
Example 1.2-26 
( )− ⋅ − +2 6 5 50 = ( ) ( )+ ⋅ − ⋅2 6 5 2 6 50 = 2 6 5 2 6 50⋅ − ⋅ = 2 30 2 300− = 2 30 2 3 100− ⋅ 
 
= 2 30 2 10 32− ⋅ = ( )2 30 2 10 3− ⋅ = 2 30 20 3− = ( )2 30 10 3− 
Example 1.2-27 
( )3 2 3 6⋅ + = ( ) ( )2 3 3 3 6⋅ + ⋅ = ( ) ( )2 3 3 3 6⋅ + ⋅ = 2 3 3 181 1⋅ + = 2 3 9 21 1+ + ⋅ 
 
= 2 3 3 22 2+ ⋅ = ( )2 3 3 2⋅ + = 6 3 2+ = ( )3 2 2+ 
Example 1.2-28 
( )3 2 10 4 20⋅ + = ( ) ( )3 2 10 3 2 4 20⋅ + ⋅ = ( ) ( )3 2 10 3 4 2 20⋅ + ⋅ ⋅ = 3 20 12 40+ 
 
= 3 4 5 12 4 10⋅ + ⋅ = 3 2 5 12 2 102 2⋅ + ⋅ = ( ) ( )3 2 5 12 2 10⋅ + ⋅ = 6 5 24 10+ = ( )6 5 4 10+ 
Example 1.2-29 
( )5 25 2163 3 3⋅ − = 5 5 63 23 33⋅ −  = 5 5 5 63 23 3 33⋅  − ⋅  = ( )5 5 5 623 3⋅  − ⋅ = 5 5 6 51 23 3⋅ − 
 
= 5 6 51 23 3+ − = 5 6 533 3− = 5 6 53− 
 
Practice Problems - Multiplying Monomial and Binomial Expressions in Radical Form, with Real Numbers 
Section 1.2b Case III Practice Problems - Multiply the following radical expressions: 
 
1. ( )2 3 2 2⋅ + = 
 
2. ( )5 8 5⋅ + = 3. ( )− ⋅ −8 3 3 = 
4. 4 98 3 23⋅ −

 = 
 
5. ( )48 324 324 4 4⋅ + = 6. ( )2 5 45 814⋅ + = 
 
Mastering Algebra - Advanced Level 1.2b Operations Involving Radical Expressions 
Hamilton Education Guides 36 
Case IV Rationalizing Radical Expressions - Monomial Denominators with Real Numbers 
Radicals are divided by each other using the following general rule: 
x
y
x
y
n
n
n
= x y≥ 〉0 0, 
In section 1.2a the difference between rational and irrational numbers was discussed. We learned 
that the square root of non perfect squares, the cube root of non perfect cubes, etc. are irrational 
numbers. For example, 3 7 10 4 73 5, , , , , .etc are classified as irrational numbers. In division 
of radicals, if the denominator of a fractional radical expression is not a rational number, we 
rationalize the denominator by changing the radicand of the denominator to a perfect square, a 
perfect cube, etc. 
Simplification of radical expressions being divided requires rationalization of the denominator. 
A monomial and irrational denominator is rationalized by multiplying the numerator and the 
denominator by the irrational denominator. This change the radicand of the denominator to a 
perfect square. 
Examples: 
1. 1
7
1
7
7
7
1 7
7 7
7
72
= × =
⋅
⋅
= = 7
7
 
Note that 7 is an irrational number. By multiplying 7 by itself the denominator is changed 
to a rational number, i.e., 7 . 
2. 20
3
20
3
4 5
3
2 5
3
2 5
3
2 5
3
3
3
2 5 3
3 3
2 15
3
2
2
= =
⋅
=
⋅
= = × =
⋅
⋅
= = 2 15
3
 
Again, note that 3 is an irrational number. By multiplying 3 by itself the denominator is 
changed to a rational number, i.e., 3 . 
Radical expressions with monomial denominators are simplified using the following steps: 
Step 1 Change the radical expression a
b
 to a
b
 and simplify. 
Step 2 Rationalize the denominator by multiplying the numerator and the denominator of the 
radical expression a
b
 by b . 
Step 3 Simplify the radical expression (see Section 1.2a, Case III). 
Examples with Steps 
The following examples show the steps as to how radical expressions with monomial 
denominators are simplified: 
Example 1.2-30 
 −8 3
32 45
 = 
Mastering Algebra - Advanced Level 1.2b Operations Involving Radical Expressions 
Hamilton Education Guides 37 
Solution: 
 Step 1 −8 3
32 45
 = − /
/ /
8
1
3
32
4
45
 = −
⋅
3
4 9 5
 = −
⋅
3
4 3 52
 = −
⋅
3
4 3 5
 = − 3
12 5
 
 
 Step 2 − 3
12 5
 = − ×3
12 5
5
5
 
 
 Step 3 − ×3
12 5
5
5
 = − ×
×
3 5
12 5 5
 = − ⋅
⋅
3 5
12 5 5
 = −
⋅
15
12 5 51 1
 = −
+
15
12 51 1
 
 
 = −
15
12 52
 = −
⋅
15
12 5
 = −
15
60 
Example 1.2-31 
 3 8
81
5
5 = 
Solution: 
 Step 1 3 8
81
5
5 = 
3 8
3
5
45
 
 
 Step 2 3 8
3
5
45
 = 3 8
3
3
3
5
45
15
15
× 
 Note that radical expressions with third, fourth, or higher root in the denominator can also 
 be rationalized by changing the denominator to a perfect third, fourth, or higher power. 
 Step 3 3 8
3
3
3
5
45
15
15
× = 3 8 3
3 3
5 15
45 15
×
×
 = 3 8 3
3 3
5
4 15
⋅
⋅
 = 3 24
3
5
4 15 +
 = 3 24
3
5
55
 = /⋅
/
3
1
24
3
1
5
 = 1 24
1
5⋅ 
 
 = 245 
 
Additional Examples: Rationalizing Radical Expressions - Monomial Denominators with Real Numbers 
The following examples further illustrate how to solve radical expressions with monomial 
denominators: 
Example 1.2-32 
8 3
2
 = 8 3
2
2
2
× = 8 3 2
2 2
×
×
 = 8 3 2
2 2
⋅
⋅
 = 8 6
2 21 1⋅
 = 8 6
21 1+
 = 8 6
22
 = /
/
8
4
6
2
1
 = 4 6
1
 = 4 6 
 
Mastering Algebra - Advanced Level 1.2b Operations Involving Radical Expressions 
Hamilton Education Guides 38 
Example 1.2-33 
5
7
2 = 5
7
 = 5
7
 = 5
7
7
7
× = 5 7
7 7
×
×
 = 5 7
7 7
⋅
⋅
 = 35
7 71 1⋅
 = 35
71 1+
 = 35
72
 = 357 
Example 1.2-34 
1
52
 = 1
5
 = 1
5
5
5
× = 1 5
5 5
×
×
 = 5
5 5⋅
 = 5
5 51 1⋅
 = 5
51 1+
 = 5
52
 = 55 
Example 1.2-35 
40 12
5 6
 = / /
/
40
8
12
5
1
6
 = 8 12
6
 = 8 4 3
6
⋅ = 8 2 3
6
2 ⋅ = ( )8 2 3
6
⋅
 = 16 3
6
 = 16 3
6
6
6
× = 16 3 6
6 6
×
×
 
 
= 16 3 6
6 6
⋅
⋅
 = 16 18
6 61 1⋅
 = 16 9 2
61 1
⋅
+
 = 16 3 2
6
2
2
⋅ = ( )16 3 2
6
⋅
 = / /
/
48
8
2
6
1
 = 8 2
1
 = 8 2 
Example 1.2-36 
1000
36
 = / / / /
/ /
1000
250
36
9
 = 250
9
 = 250
9
 = 25 10
32
⋅ = 5 10
3
2 × = 5 103 
Practice Problems: Rationalizing Radical Expressions - Monomial Denominators with Real Numbers 
Section 1.2b Case IV Practice Problems - Solve the following radical expressions: 
1. 1
8
 = 
 
2. 50
7
2 = 3. 75
5−
 = 
4. 25
16
3 = 
 
5. 32
8
5 = 6. −
−
3 100
5 3000
 = 
 
Mastering Algebra - Advanced Level 1.2b Operations Involving Radical Expressions 
Hamilton Education Guides 39 
Case V Rationalizing Radical Expressions - Binomial Denominators with Real Numbers 
Simplification of fractional radical expressions with binomial denominators requires rationalization 
of the denominator. A binomial denominator is rationalized by multiplying the numerator and the 
denominator by its conjugate. Two binomials that differ only by the sign between them are called 
conjugates of each other. Note that whenever conjugates are multiplied by each other, the two 
similar but opposite in sign middle terms drop out. 
Examples: 
1. The conjugate of 2 3+ is 2 3− . 
2. The conjugate of 6 10− is 6 10+ . 
3. The conjugate of 3 5− is 3 5+ . 
4. The conjugate of 7 2+ is 7 2− . 
Radical expressions with binomial denominators are simplified using the following steps: 
Step 1 Simplify the radical terms in the numerator and the denominator (see Section 1.2a, 
Case III). 
Step 2 Rationalize the denominator by multiplying the numerator and the denominator by its 
conjugate. 
Step 3 Simplify the radical expression using the FOIL method (see Section 1.2b, Case II). 
Examples with Steps 
The following examples show the steps as to how radical expressions with two terms in the 
denominator are simplified: 
Example 1.2-37 
 8
2 2−
 
Solution: 
 Step 1 Not Applicable 
 
 Step 2 8
2 2−
 = 8
2 2
2 2
2 2−
×
+
+
 
 
 Step 3 8
2 2
2 2
2 2−
×
+
+
 = 
( )
( ) ( )
8 2 2
2 2 2 2
× +
− × +
 = 
( )
( ) ( ) ( ) ( )
8 2 2
2 2 2 2 2 2 2 2
⋅ +
⋅ + ⋅− ⋅ − ⋅
 
 
 = 
( )8 2 2
4 2 2 2 2 2 2
+
+ − − ⋅
 = 
( )8 2 2
4 2 21 1
+
− ⋅
 = 
( )8 2 2
4 21 1
+
− +
 = 
( )8 2 2
4 22
+
−
 = 
( )8 2 2
4 2
+
−
 
 
Mastering Algebra - Advanced Level 1.2b Operations Involving Radical Expressions 
Hamilton Education Guides 40 
 = 
( )/ +
/
8
4
2 2
2
 = 
( )4 2 2
1
+
 = ( )4 2 2+ 
Example 1.2-38 
 8 4
4 2
+
−
 = 
Solution: 
 Step 1 8 4
4 2
+
−
 = 2 2 2
4 2
2 2⋅ +
−
 = 2 2 2
4 2
+
−
 
 
 Step 2 2 2 2
4 2
+
−
 = 2 2 2
4 2
4 2
4 2
+
−
×
+
+
 
 
 Step 3 2 2 2
4 2
4 2
4 2
+
−
×
+
+
 = 
( ) ( )
( ) ( )
2 2 2 4 2
4 2 4 2
+ × +
− × +
 = 
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
4 2 2 2 2 2 2 4 2 2
4 4 4 2 4 2 2 2
⋅ + ⋅ + ⋅ + ⋅
⋅ + ⋅ − ⋅ − ⋅
 
 
 = 8 2 2 2 2 8 2 2
16 4 2 4 2 2 2
+ ⋅ + +
+ − − ⋅
 = 8 2 2 2 8 2 2
16 2
2
2
+ + +
−
 = ( )8 2 2 2 8 2 2
16 2
+ ⋅ + +
−
 
 
 = 8 2 4 8 2 2
14
+ + + = ( )8 2 2 12
14
+ +
 = 10 2 12
14
+ = 
( )/⋅ +
/ /
2
1
5 2 6
14
7
 = 
( )1 5 2 6
7
⋅ +
 
 = 5 2 67
+
 
Additional Examples: Rationalizing Radical Expressions - Binomial Denominators with Real Numbers 
The following examples further illustrate how to rationalize radical expressions with binomial 
denominators: 
Example 1.2-39 
5
3 5+
 = 5
3 5
3 5
3 5+
×
−
−
 = 
( )
( ) ( )
5 3 5
3 5 3 5
× −
+ × −
 = 
( ) ( )
( ) ( ) ( ) ( )
3 5 5 5
3 3 3 5 3 5 5 5
⋅ − ⋅
⋅ − ⋅ + ⋅ − ⋅
 
 
= 3 5 5
9 3 5 3 5 5
2
2
−
− + −
 = 3 5 5
9 52
−
−
 = 3 5 5
9 5
−
−
 = 3 5 54
− 
Example 1.2-40 
5 3
7 2
+
+
 = 5 3
7 2
7 2
7 2
+
+
×
−
−
 = 
( ) ( )
( ) ( )
5 3 7 2
7 2 7 2
+ × −
+ × −
 = 
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
5 7 5 2 3 7 3 2
7 7 7 2 2 7 2 2
⋅ − ⋅ + ⋅ − ⋅
⋅ − ⋅ + ⋅ − ⋅
 
Mastering Algebra - Advanced Level 1.2b Operations Involving Radical Expressions 
Hamilton Education Guides 41 
= 5 7 5 2 3 7 3 2
7 7 7 2 2 7 2 2
⋅ − ⋅ + ⋅ − ⋅
⋅ − ⋅ + ⋅ − ⋅
 = 35 10 21 6
7 14 14 22 2
− + −
− + −
 = 35 10 21 6
7 2
− + −
−
 
 
= 35 10 21 65
− + − 
Example 1.2-41 
3 27
3 3
+
−
 = 3 9 3
3 3
+ ⋅
−
 = 3 3 3
3 3
2+ ⋅
−
 = 3 3 3
3 3
+
−
 = 3 3 3
3 3
3 3
3 3
+
−
×
+
+
 = 
( ) ( )
( ) ( )
3 3 3 3 3
3 3 3 3
+ × +
− × +
 
 
= 
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
3 3 3 3 3 3 3 3 3 3
3 3 3 3 3 3 3 3
⋅ + ⋅ + ⋅ + ⋅
⋅ + ⋅ − ⋅ − ⋅
 = 9 3 3 9 3 3 3 3
9 3 3 3 3 3 3
+ + + ⋅
+ − − ⋅
 = 9 3 3 9 3 3 3
9 3
2
2
+ + +
−
 
 
= 9 3 3 9 3 3 3
9 3
+ + + ⋅
−
 = ( ) ( )9 9 3 9 3
9 3
+ + +
−
 = 18 12 3
6
+ = 
( )/ +
/
6 3 2 3
6
 = 3 2 3
1
+ = 3 2 3+ 
Example 1.2-42 
5 1
5 1
−
+
 = 
5 1
5 1
5 1
5 1
−
+
×
−
−
 = 
( ) ( )
( ) ( )
5 1 5 1
5 1 5 1
− × −
+ × −
 = 
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
5 5 1 5 1 5 1 1
5 5 1 5 1 5 1 1
⋅ − ⋅ − ⋅ + ⋅
⋅ − ⋅ + ⋅ − ⋅
 = 
5 5 5 5 1
5 5 5 5 1
⋅ − − +
⋅ − + −
 
 
= 5 5 5 1
5 1
2
2
− − +
−
 = 5 5 5 1
5 1
− − +
−
 = 
( ) ( )5 1 1 1 5
5 1
+ + − −
−
 = 6 2 5
4
− = 
( )/ −
/
2 3 5
4
2
 = 
3 5
2
−
 
Practice Problems: Rationalizing Radical Expressions - Binomial Denominators with Real Numbers 
Section 1.2b Case V Practice Problems - Solve the following radical expressions: 
1. 7
1 7+
 = 
 
2. 1 18
2 18
−
+
 = 3. 5
5 2+
 = 
4. 3 5
7 4
−
−
 = 
 
5. − +
+
3 3
4 5
 = 6. 3 3
3 3
−
+
 = 
 
Mastering Algebra - Advanced Level 1.2b Operations Involving Radical Expressions 
Hamilton Education Guides 42 
Case VI Adding and Subtracting Radical Terms 
Radicals are added and subtracted using the following general rule: 
( )k a k a k a k k k an n n n1 2 3 1 2 3+ + = + + 
Only similar radicals can be added and subtracted. Similar radicals are defined as radical 
expressions with the same index n and the same radicand a . Note that the distributive property 
of multiplication is used to group the numbers in front of the similar radical terms. Radicals are 
added and subtracted using the following steps: 
Step 1 Group similar radicals. 
Step 2 Simplify the radical expression. 
Examples with Steps 
The following examples show the steps as to how radical expressions are added and subtracted: 
Example 1.2-43 
 6 2 4 2+ = 
Solution: 
 Step 1 6 2 4 2+ = ( )6 4 2+ 
 
 Step 2 ( )6 4 2+ = 10 2 
 
Example 1.2-44 
 20 3 8 3 5 35 5 5− + = 
Solution: 
 Step 1 20 3 8 3 5 35 5 5− + = ( )20 8 5 35− + 
 
 Step 2 ( )20 8 5 35− + = 17 35 
 
Example 1.2-45 
 ( )6 7 2 7 2 73+ − = 
Solution: 
 Step 1 ( )6 7 2 7 2 73+ − = ( )6 2 7 2 73+ − 
 
 Step 2 ( )6 2 7 2 73+ − = 8 7 2 73− 
 
 
 
 
Mastering Algebra - Advanced Level 1.2b Operations Involving Radical Expressions 
Hamilton Education Guides 43 
Additional Examples - Adding and Subtracting Radical Terms 
The following examples further illustrate how to add and subtract radical terms: 
Example 1.2-46 
2 5 3 5 6+ + = ( )2 3 5 6+ + = 5 5 6+ 
Example 1.2-47 
8 4 2 4 53 3+ + = ( )8 2 4 53+ + = 10 4 53 + = ( )5 2 4 13 + 
Example 1.2-48 
2 3 4 3 3 3 54 4 4 4+ − + = ( )2 4 3 3 54 4+ − + = 3 3 54 4+ 
Note that the two radical terms have the same index (4) but have different radicands (3 and 
5). Therefore, they can not be combined. 
Example 1.2-49 
( )5 3 5 5 4 25 5 5+ + − +a a = ( ) ( )1 3 5 4 25+ + − +a a = ( ) ( )4 5 4 25+ − +a a = ( )[ ]4 5 25+ −a 
Example 1.2-50 
5 2 8 2 2 2 4 2 8 23 3 3x x c x x x+ − + − = ( ) ( )5 8 2 2 4 8 23+ − + −c x x = ( )13 2 2 4 23− −c x x 
Example 1.2-51 
a xy b xy c xy d+ − −3 2 = a xy c xy b xy d− + −2 3 = ( )a c xy b xy d− + −2 3 
Example 1.2-52 
2 75 3 125 20 3 10 4 10+ + + − = ( )2 25 3 3 25 5 4 5 3 4 10⋅ + ⋅ + ⋅ + − 
 
= 2 5 3 3 5 5 2 5 102 2 2⋅ + ⋅ + ⋅ − = ( ) ( )2 5 3 3 5 5 2 5 10⋅ + ⋅ + − = 10 3 15 5 2 5 10+ + − 
 
= ( )10 3 15 2 5 10+ + − = 10 3 17 5 10+ − 
Example 1.2-53 
8 6 4 6 6 5 4 53 3 3+ + − −a = ( ) ( )8 4 6 1 4 53+ + + − −a = ( ) ( )12 6 5 53+ + −a = ( )12 6 5 53+ −a 
 
Practice Problems - Adding and Subtracting Radical Terms 
 
Section 1.2b Case VI Practice Problems - Simplify the following radical expressions: 
1. 5 3 8 3+ = 
 
2. 2 3 4 33 3− = 3. 12 5 8 5 2 34 4 4+ + = 
4. a ab b ab c ab− + = 
 
5. 3 2 43 3 23x x x x x x− + = 6. 5 2 8 53 3+ = 
 
Mastering Algebra - Advanced Level 1.3a Factoring Polynomials Using the Greatest Common Factoring Method 
Hamilton Education Guides 44 
1.3 Factoring Polynomials 
 
In this section different methods for factoring polynomials are reviewed. The steps in factoring 
polynomials using the Greatest Common Factoring, the Grouping, and the Trial and Error 
method are addressed in Sections 1.3a, 1.3b, and 1.3c, respectively. Other factoring methods 
such as using the Difference of Two Square method, the Sum and Difference of Two Cubes 
method, as well as the method for factoring Perfect Square Trinomials are discussed in Section 
1.3d. 
1.3a Factoring Polynomials Using the Greatest Common Factoring Method 
Solving algebraic fractions, which are reviewed in Section 1.5, requires a thorough knowledge of 
the factoring and solution methods that are introduced in this and the following section. 
Therefore, it is essential that students learn how to factor polynomials of second or higher 
degrees. For example, simplification of math operations such as: 
x y
x y
x xy y
y y x
−
−
⋅
+ −
− +2
2
2 3
2 2
2 2 ; 
x y
x xy y
x xy y
x y
2 2
2 2
2 2
2 5 3
4 4 3
5 5
−
+ +
⋅
+ −
−
 ; x
x x x
x
x
2
3 2
29
5 6
16
8 24
−
− +
⋅
+
 
require familiarization with various factoring methods. In this section students are introduced to 
factoring the Greatest Common Factor to: monomial terms (Case I) and binomial and 
polynomial terms (Case II). 
Case I Factoring the Greatest Common Factor to Monomial Terms 
Factoring a polynomial means writing the polynomial as a product of two or more simpler 
polynomials. One method in factoring polynomials is by using the Greatest Common Factoring 
method where the Greatest Common Factor (G.C.F.) is factored out. The Greatest Common 
Factor to monomial terms is foundusing the following steps: 
Step 1 a. Write the numbers and the variables in their prime factored form. 
 b. Identify the prime numbers and variables that are common in monomials. 
Step 2 Multiply the common prime numbers and variables to obtain the G.C.F. 
The following examples show the steps as to how monomial expressions are factored using the 
Greatest Common Factoring method: 
Example 1.3-1 
 Find the G.C.F. to 27 2 3x y , 9 3x y , and 15 2xy . 
Solution: 
 Step 1 27 2 3x y = 3 9 2⋅ ⋅ ⋅ ⋅ ⋅x x y y = 3 3 3⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅x x y y y 
 
 9 3x y = 3 3 2⋅ ⋅ ⋅ ⋅x x y = 3 3⋅ ⋅ ⋅ ⋅ ⋅x x x y and 
 
 15 2xy = 3 5⋅ ⋅ ⋅ ⋅x y y 
 
 Therefore, the common terms are 3 , x and y . 
Mastering Algebra - Advanced Level 1.3a Factoring Polynomials Using the Greatest Common Factoring Method 
Hamilton Education Guides 45 
 
 Step 2 G.C.F. = 3⋅ ⋅x y = 3xy 
Example 1.3-2 
 Find the G.C.F. to 32 3 3a b , 46 2ab , and 56 2 4a b . 
Solution: 
 Step 1 32 3 3a b = 4 8 2 2⋅ ⋅ ⋅ ⋅ ⋅a a b b = 2 2 2 4⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅a a a b b b = 2 2 2 2 2⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅a a a b b b 
 
 46 2ab = 3 12⋅ ⋅ ⋅ ⋅a b b = 3 3 4⋅ ⋅ ⋅ ⋅ ⋅a b b = 3 3 2 2⋅ ⋅ ⋅ ⋅ ⋅ ⋅a b b and 
 
 56 2 4a b = 7 8 2 2⋅ ⋅ ⋅ ⋅ ⋅a a b b = 7 2 4⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅a a b b b b = 7 2 2 2⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅a a b b b b 
 
 Therefore, the common terms are 2 , 2 , a , b , and b . 
 
 Step 2 G.C.F. = 2 2⋅ ⋅ ⋅ ⋅a b b = 4 2ab 
 
Additional Examples - Factoring the Greatest Common Factor to Monomial Terms 
The following examples further illustrate how to find the Greatest Common Factor to monomial 
terms: 
Example 1.3-3 
Find the G.C.F. to 48 2xy , 16 2x y , 4 3 2x y , and 12xy . 
 
1. 48 2xy = 12 4⋅ ⋅ ⋅ ⋅x y y = 6 2 2 2⋅ ⋅ ⋅ ⋅ ⋅ ⋅x y y = 3 2 2 2 2⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅x y y 
 
2. 16 2x y = 2 8⋅ ⋅ ⋅ ⋅x x y = 2 2 4⋅ ⋅ ⋅ ⋅ ⋅x x y = 2 2 2 2⋅ ⋅ ⋅ ⋅ ⋅ ⋅x x y 
 
3. 4 3 2x y = 2 2 2⋅ ⋅ ⋅ ⋅ ⋅x x y y = 2 2⋅ ⋅ ⋅ ⋅ ⋅ ⋅x x x y y 
 
4. 12xy = 3 4⋅ ⋅ ⋅x y = 3 2 2⋅ ⋅ ⋅ ⋅x y 
 
Therefore, the common terms are 2 , 2 , x , and y . Thus, G.C.F. = 2 2⋅ ⋅ ⋅x y = 4xy 
Example 1.3-4 
Find the G.C.F. to 55 2 3u w z , 50 2 2uw z , and 15 3u w . 
 
1. 55 2 3u w z = 5 11 2⋅ ⋅ ⋅ ⋅ ⋅ ⋅u u w w z = 5 11⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅u u w w w z 
 
2. 50 2 2uw z = 5 10⋅ ⋅ ⋅ ⋅ ⋅ ⋅u w w z z = 5 5 2⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅u w w z z 
 
3. 15 3u w = 5 3 2⋅ ⋅ ⋅ ⋅u u w = 5 3⋅ ⋅ ⋅ ⋅ ⋅u u u w 
 
Therefore, the common terms are 5 , u , and w . Thus, G.C.F. = 5 ⋅ ⋅u w = 5uw 
 
Mastering Algebra - Advanced Level 1.3a Factoring Polynomials Using the Greatest Common Factoring Method 
Hamilton Education Guides 46 
 
Example 1.3-5 
Find the G.C.F. to 27abc , 36 2 2 3a b c , and 24 2ac . 
 
1. 27abc = 9 3⋅ ⋅ ⋅ ⋅a b c = 3 3 3⋅ ⋅ ⋅ ⋅ ⋅a b c 
 
2. 36 2 2 3a b c = 2 18 2⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅a a b b c c = 2 2 9⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅a a b b c c c = 2 2 3 3⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅a a b b c c c 
 
3. 24 2ac = 3 8⋅ ⋅ ⋅ ⋅a c c = 3 2 4⋅ ⋅ ⋅ ⋅ ⋅a c c = 3 2 2 2⋅ ⋅ ⋅ ⋅ ⋅ ⋅a c c 
 
Therefore, the common terms are 3 , a , and c . Thus, G.C.F. = 3⋅ ⋅a c = 3ac 
Example 1.3-6 
Find the G.C.F. to 12x , 60 2xy , and 63 2x . 
 
1. 12x = 3 4⋅ ⋅ x = 3 2 2⋅ ⋅ ⋅ x 
 
2. 60 2xy = 4 15⋅ ⋅ ⋅ ⋅x y y = 2 2 5 3⋅ ⋅ ⋅ ⋅ ⋅ ⋅x y y 
 
3. 63 2x = 3 21⋅ ⋅ ⋅x x = 3 3 7⋅ ⋅ ⋅ ⋅x x 
 
Therefore, the common terms are 3 and x . Thus, G.C.F. = 3⋅ x = 3x 
 
Practice Problems - Factoring the Greatest Common Factor to Monomial Terms 
Section 1.3a Case I Practice Problems - Find the Greatest Common Factor to the following 
monomial terms: 
1. 5 3x and 15x 
 
2. 18 2 3 4x y z and 24 4 5xy z 3. 16 2 3a bc , 38 4 2ab c , and 6 3a bc 
4. r s5 4 , 4 3 2r s , and 3rs 
 
5. 10 2 3u vw , 2 3 2uv w , and uv2 6. 19 3 3a b , 12 2ab , and 6ab 
 
 
Mastering Algebra - Advanced Level 1.3a Factoring Polynomials Using the Greatest Common Factoring Method 
Hamilton Education Guides 47 
Case II Factoring the Greatest Common Factor to Binomial and Polynomial Terms 
The concept of obtaining Greatest Common Factor can be extended to binomial expressions by 
obtaining the greater common monomial factor which is found by using the following steps: 
Step 1 a. Write each monomial term in its prime factored form. 
 b. Identify the prime numbers and variables that are common to monomials. 
 c. Multiply the common prime numbers and variables to obtain the greatest common 
monomial factor. 
Step 2 Factor out the greatest common monomial factor from the binomial expression. 
The following examples show the steps as to how binomial expressions are factored: 
Example 1.3-7 
 Factor 6 23 2 2 2 2a b c a bc− . 
Solution: 
 Step 1 6 3 2 2a b c = 2 3⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅a a a b b c c and 
 
 2 2 2a bc = 2 ⋅ ⋅ ⋅ ⋅ ⋅a a b c c 
 
 Therefore, the common terms are 2 , a , a , b , c , and c which implies that 
 the greatest common monomial factor is 2 2 2 2⋅ ⋅ ⋅ ⋅ ⋅ =a a b c c a bc . Thus, 
 
 Step 2 6 23 2 2 2 2a b c a bc− = ( )2 3 12 2a bc ab − 
Example 1.3-8 
 Factor 12 363 2 2 2x y z x z+ . 
Solution: 
 Step 1 12 3 2x y z = 4 3 2⋅ ⋅ ⋅ ⋅ ⋅ ⋅x x y y z = 2 2 3⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅x x x y y z and 
 
 36 2 2x z = 2 18⋅ ⋅ ⋅ ⋅ ⋅x x z z = 2 2 9⋅ ⋅ ⋅ ⋅ ⋅ ⋅x x z z = 2 2 3 3⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅x x z z 
 
 Therefore, the common terms are 2 , 2 , 3 , x , x , and z which implies that 
 the greatest common monomial factor is 2 2 3 12 2⋅ ⋅ ⋅ ⋅ ⋅ =x x z x z . Thus, 
 
 Step 2 12 363 2 2 2x y z x z+ = ( )12 32 2x z xy z+ 
 
Additional Examples - Factoring the Greatest Common Factor to Binomial and Polynomial Terms 
The following examples further illustrate how to find the greatest common monomial factor to 
binomial terms: 
Example 1.3-9 
Find the greatest common monomial factor to 35 52 3 2m n mn+ . 
Mastering Algebra - Advanced Level 1.3a Factoring Polynomials Using the Greatest Common Factoring Method 
Hamilton Education Guides 48 
1. 35 2 3m n = 5 7 2⋅ ⋅ ⋅ ⋅ ⋅m m n n = 5 7⋅ ⋅ ⋅ ⋅ ⋅ ⋅m m n n n 
 
2. 5 2mn = 5 ⋅ ⋅ ⋅m n n 
 
Therefore, the common terms are 5 , m , n , and n which implies that the greatest common 
monomial factor is 5 5 2⋅ ⋅ ⋅ =m n n mn . Thus, 35 52 3 2m n mn+ = ( )5 7 12mn mn + 
Example 1.3-10 
Find the greatest common monomial factor to 6 662 4a b ab+ . 
 
1. 6 2a b = 2 3⋅ ⋅ ⋅ ⋅a a b 
 
2. 66 4ab = 6 11 2 2⋅ ⋅ ⋅ ⋅a b b = 2 3 11⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅a b b b b 
 
Therefore, the common terms are 2 , 3 , a , and b which implies that the greatest common 
monomial factor is 2 3 6⋅ ⋅ ⋅ =a b ab . Thus, 6 662 4a b ab+ = ( )6 11 3ab a b+ 
Example 1.3-11 
Find the greatest common monomial factor to 7 493 6 2 5p q p q− . 
 
1. 7 3 6p q = 7 2 3 3⋅ ⋅ ⋅ ⋅p p q q = 7 2 2⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅p p p q q q q = 7 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅p p p q q q q q q 
 
2. 49 2 5p q = 7 7 4⋅ ⋅ ⋅ ⋅ ⋅p p q q = 7 7 2 2⋅ ⋅ ⋅ ⋅ ⋅ ⋅p p q q q = 7 7⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅p p q q q q q 
 
Therefore, the common terms are 7 , p , p , q , q , q , q , and q which implies that the greatest 
common monomial factor is 7 7 2 5⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ =p p q q q q q p q . Thus, 7 493 6 2 5p q p q− = ( )7 72 5p q pq − 
Example 1.3-12 
Find the greatest common monomial factor to 48 20x xy− . 
 
1. 48x = 4 12⋅ ⋅ x = 2 2 4 3⋅ ⋅ ⋅ ⋅ x = 2 2 2 2 3⋅ ⋅ ⋅ ⋅ ⋅ x 
 
2. 20xy = 4 5⋅ ⋅ ⋅x y = 2 2 5⋅ ⋅ ⋅ ⋅x y 
 
Therefore, the common terms are 2 , 2 , and x which implies that the greatest common 
monomial factor is 2 2 4⋅ ⋅ =x x . Thus, 48 20x xy− = ( )4 12 5x y− 
Example 1.3-13 
Find the greatest common monomial factor to 24 123 3 2 4x y x y+ . 
 
1. 24 3 3x y = 2 12 2 2⋅ ⋅ ⋅ ⋅ ⋅x x y y = 2 4 3⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅x x x y y y = 2 2 2 3⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅x x x y y y 
 
2. 12 2 4x y = 2 6 2 2⋅ ⋅ ⋅ ⋅ ⋅x x y y = 2 2 3⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅x x y y y y 
Mastering Algebra - Advanced Level 1.3a Factoring Polynomials Using the Greatest Common Factoring Method 
Hamilton Education Guides 49 
Therefore, the common terms are 2 , 2 , 3 , x , x , y , y and y which implies that the greatest 
common monomial factor is 2 2 3 12 2 3⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ =x x y y y x y . Thus, 24 123 3 2 4x y x y+= ( )12 22 3x y x y+ 
Note 1: As one gains more proficiency in solving this class of problems the need for factoring 
each monomial term to its prime factored form lessens. Therefore, students may simplify a given 
binomial expression by mentally factoring out the common terms. For example, we can quickly 
factor out the expression 24 124 2 2 3 2x y z x y z+ by observing that its greatest common monomial 
factor is 12 2 2x y z , e.g., 24 124 2 2 3 2x y z x y z+ = ( )12 22 2 2x y z x yz+ . 
Note 2: The process of factoring binomial expressions can further be expanded to include 
trinomials and polynomials. Following are few additional examples indicating how the greatest 
common monomial factor to polynomials is obtained: 
1. 5 15 502 3x x x+ + = ( )5 1 3 10 2x x x+ + 2. 24 15 122xy xy y+ + = ( )3 8 5 4y xy x+ + 
3. 8 4 22 3 2 2a b ab a b+ − = ( )2 4 22ab ab b a+ − 4. 20 15 52 2 3 2u w u w uw+ + = ( )5 4 3 12uw uw u w+ + 
5. 4 6 23 2x x x+ + = ( )2 2 3 12x x x+ + 6. 20 5 152 2 3x y x y y− + = ( )5 4 32 3y x y x− + 
7. 40 12 83 4 2 2 2 2 2x y z x y z x yz− + = ( )4 10 3 22 3x yz xy z yz− + 
Practice Problems - Factoring the Greatest Common Factor to Binomial and Polynomial Terms 
Section 1.3a Case II Practice Problems - Find the Greatest Common Factor to the following 
binomial and polynomial terms: 
 
1. 18 123 3 2x y x y− = 
 
2. 3 152 3 2 3a b c ab c+ = 3. xyz x y z3 2 2 54+ = 
4. 25 53 2 3p p q pq+ + = 
 
5. r s t rst2 2 25− = 6. 36 4 123 3 2 4 3 3x yz xy z x y z+ − = 
 
Mastering Algebra - Advanced Level 1.3b Factoring Polynomials Using the Grouping Method 
Hamilton Education Guides 50 
1.3b Factoring Polynomials Using the Grouping Method 
In many instances polynomials with four or more terms have common terms that can be grouped 
together. This process is called factoring by grouping. The steps as to how polynomials are 
grouped are shown below: 
Step 1 Factor the common variables, or numbers, from the monomial terms. 
Step 2 Factor the common binomial factor obtained in step 1 by grouping. 
Examples with Steps 
The following examples show the steps as to how polynomials are factored using the grouping 
method: 
Example 1.3-14 
 u u u2 2 16 32− − + = 
Solution: 
 Step 1 u u u2 2 16 32− − + = ( ) ( )u u u− − −2 16 2 
 
 Step 2 ( ) ( )u u u− − −2 16 2 = ( )( )u u− −2 16 
Example 1.3-15 
 60 24 15 62m m m+ − − = 
Solution: 
 Step 1 60 24 15 62m m m+ − − = ( ) ( )12 5 2 3 5 2m m m+ − + 
 
 Step 2 ( ) ( )12 5 2 3 5 2m m m+ − + = ( )( )5 2 12 3m m+ − = ( )( )4 5 2 3 1m m+ − 
Example 1.3-16 
 15 25 9 153 2y y y+ + + = 
Solution: 
 Step 1 15 25 9 152y y y+ + + = ( ) ( )5 3 5 3 3 5y y y+ + + 
 
 Step 2 ( ) ( )5 3 5 3 3 5y y y+ + + = ( )( )3 5 5 3y y+ + 
 
Additional Examples - Factoring Polynomials Using the Grouping Method 
The following examples further illustrate how to factor polynomials using the grouping method: 
Example 1.3-17 
( )5 15 152x y x y+ + + = ( ) ( )5 152x y x y+ + + = ( ) ( )[ ]5 3x y x y+ + + 
Example 1.3-18 
( ) ( ) ( )8 4 22 3a b a b a b+ + + + + = ( ) ( ) ( )[ ]2 4 2 12a b a b a b+ + + + + = ( ) ( ) ( )[ ]{ }2 2 2 1a b a b a b+ + + + + 
Mastering Algebra - Advanced Level 1.3b Factoring Polynomials Using the Grouping Method 
Hamilton Education Guides 51 
Example 1.3-19 
− + − +x x x2 2 3 6 = ( ) ( )x x x− + + − +2 3 2 = ( )( )− + +x x2 3 
Example 1.3-20 
3 7 3 7ab b a− − + = ( ) ( )b a a3 7 3 7− − + = ( )( )3 7 1a b− − 
Example 1.3-21 
x x x3 25 5+ + + = ( )x x x2 5 5+ + + = ( ) ( )x x x2 5 5+ + + = ( )( )x x+ +5 12 
Example 1.3-22 
y y y3 23 4 12+ + + = ( ) ( )y y y2 3 4 3+ + + = ( )( )y y+ +3 42 
Example 1.3-23 
24 12 36 182x y xy x− − + = ( ) ( )12 2 1 18 2 1xy x x− − − = ( )( )2 1 12 18x xy− − = ( )( )6 2 1 2 3x xy− − 
Example 1.3-24 
12 6 4 23 2r s r s r− + − = ( ) ( )6 2 1 2 2 12r s r r− + − = ( )( )2 1 6 22r r s− + = ( )( )2 2 1 3 12r r s− + 
In the following sections, additional factoring methods are reviewed. These methods are used to 
present polynomials in their equivalent factored form. Students are encouraged to spend 
adequate time learning each method. 
Practice Problems - Factoring Polynomials Using the Grouping Method 
Section 1.3b Practice Problems - Factor the following polynomials using the grouping method: 
 
1. 2 5 6 15ab b a− − + = 
 
2. y y y3 24 4+ + + = 3. 42 21 70 352x y xy x+ − − = 
4. ( ) ( )x y x y x y+ + + + +3 2 = 
 
5. ( )4 32 322a b a b+ + + = 6. 36 6 18 33 2r s r s r− + − = 
 
 
Mastering Algebra - Advanced Level 1.3c Factoring Polynomials Using the Trial and Error Method 
Hamilton Education Guides 52 
1.3c Factoring Polynomials Using the Trial and Error Method 
Expressing trinomials as the product of two binomials is one of the most common ways of 
factoring. In this section, we will review how to factor trinomials of the form ax bx c2 + + , where 
a = 1 (Case I) and where 1〉a (Case II), using a factoring method which in this book is referred to 
as the Trial and Error method. 
 
Case I Factoring Trinomials of the Form ax bx c2 + + where a = 1 
To express a trinomial of the form ax bx c2 + + , where a = 1, in its factored form ( ) ( )x m x n+ + , let 
us consider the product ( ) ( )x m x n+ + and use the FOIL method to see how each term of the 
resulting trinomial is formed , i.e., 
( ) ( )x m x n+ + = x x n x m x m n⋅ + ⋅ + ⋅ + ⋅ = ( )x m n x mn2 + + + 
Note that the coefficient of the x term is the sum of m and n and the constant term is the 
product of m and n . We use this concept in order to express trinomials of the form x bx c2 + + 
in their equivalent factored form. In addition, in order to choose the right sign for the two integer 
numbers m and n , the knowledge of the following general sign rules for the indicated cases is 
needed: 
General Sign Rules 
When factoring a trinomial of the form x ax b2 + + to its equivalent factored form of ( ) ( )x m x n+ + , 
the sign of the two integer numbers m and n is determined based on the following cases: 
Case I. If the sum of the two integer numbers ( )a b+ is positive ( )+ and the product of the two 
integer numbers ( )a b⋅ is negative ( )− , then the two integer numbers m and n must have opposite 
signs. 
Case II. If the sum of the two integer numbers ( )a b+ is negative ( )− and the product of the two 
integer numbers ( )a b⋅ is positive ( )+ , then the two numbers must have the same sign. However, 
since the sum is negative, the two integer numbers m and n must both be negative. 
Case III. If the sum of the two integer numbers ( )a b+ is positive ( )+ and the product of the two 
integer numbers ( )a b⋅ is also positive ( )+ , then the two integer numbers m and n must both be 
positive. 
Case IV. If the sum of the two integer numbers ( )a b+ is negative ( )− and the product of the two 
integer numbers ( )a b⋅ is also negative ( )− , then the two integer numbers m and n must have 
opposite signs. 
General Sign Rules - Summary 
If sign of the sum ( )a b+ is and sign of the product ( )a b⋅ is then, the two integer 
numbers m and n must 
+ − have opposite signs 
− + have negative signs 
+ + have positive signs 
− − have opposite signs 
Mastering Algebra - Advanced Level 1.3c Factoring Polynomials Using the Trial and Error Method 
Hamilton Education Guides 53 
To factor a trinomial of the form x ax b2 + + to its equivalent factored form of ( ) ( )x m x n+ + use 
the following steps: 
Step 1 Obtain two numbers m and n whose sum equals to a and whose product equals to b . 
Step 2 Write the trinomial in its factored form. Check the answer by using the FOIL method. 
Examples with Steps 
The following examples show the steps as to how trinomials of the form x ax b2 + + are factored: 
Example 1.3-25 
 Factor x x2 16 55− + . 
Solution: 
 Step 1 Obtain two numbers whose sum is −16 and whose product is 55 . Notethat 
 since the sum is negative and the product is positive the two integer numbers 
 must both be negative (Case II). Let’s construct a table as follows: 
 
Sum Product 
− − = −15 1 16 ( ) ( )− ⋅ − =15 1 15 
− − = −14 2 16 ( ) ( )− ⋅ − =14 2 28 
− − = −13 3 16 ( ) ( )− ⋅ − =13 3 39 
− − = −12 4 16 ( ) ( )− ⋅ − =12 4 48 
− − = −11 5 16 ( ) ( )− ⋅ − =11 5 55 
 
 Step 2 The last line contains the sum and the product of the two numbers that we 
 need. Therefore, x x2 16 55− + = ( )( )x x− −11 5 
 Check: ( )( )x x− −11 5 = ( ) ( )x x x x⋅ − ⋅ − ⋅ + − −5 11 11 5 = ( )x x2 5 11 55+ − − + = x x2 16 55− + 
Example 1.3-26 
 Factor x x2 2 48+ − . 
Solution: 
 Step 1 Obtain two numbers whose sum is +2 and whose product is −48 . Note that 
 since the sum is positive and the product is negative the two integer numbers 
 must have opposite signs (Case I). Let’s construct a table as follows: 
Sum Product 
10 8 2− = ( )10 8 80⋅ − = − 
9 7 2− = ( )9 7 63⋅ − = − 
8 6 2− = ( )8 6 48⋅ − = − 
 
 Step 2 The last line contains the sum and the product of the two numbers that we 
 need. Therefore, x x2 2 48+ − = ( )( )x x+ −8 6 
 Check: ( )( )x x+ −8 6 = ( )x x x x⋅ − ⋅ + ⋅ + ⋅ −6 8 8 6 = ( )x x2 6 8 48+ − + − = x x2 2 48+ − 
 
Mastering Algebra - Advanced Level 1.3c Factoring Polynomials Using the Trial and Error Method 
Hamilton Education Guides 54 
Additional Examples - Factoring Trinomials of the Form ax bx c2 + + where a = 1 
The following examples further illustrate how to factor trinomials using the Trial and Error method: 
Example 1.3-27: Factor w w2 9 20+ + . 
Solution: 
To factor the above trinomial we need to obtain two numbers whose sum is 9 and whose 
product is 20 . Let’s construct a table as follows: 
Sum Product 
1 8 9+ = 1 8 8⋅ = 
2 7 9+ = 2 7 14⋅ = 
3 6 9+ = 3 6 18⋅ = 
4 5 9+ = 4 5 20⋅ = 
 
The last line contains the sum and the product of the two numbers that we need. Thus, 
w w2 9 20+ + = ( )( )w w+ +5 4 
Check: ( )( )w w+ +5 4 = w w w w⋅ + ⋅ + ⋅ + ⋅4 5 4 5 = w w w2 4 5 20+ + + = ( )w w2 4 5 20+ + + = w w2 9 20+ + 
Example 1.3-28: Factor x x2 10 12− + . 
Solution: 
To factor the above trinomial we need to obtain two numbers whose sum is −10 and whose 
product is 12 . Let’s construct a table as follows: 
Sum Product 
− − = −1 9 10 ( ) ( )− ⋅ − =1 9 9 
− − = −2 8 10 ( ) ( )− ⋅ − =2 8 16 
− − = −3 7 10 ( ) ( )− ⋅ − =3 7 21 
− − = −4 6 10 ( ) ( )− ⋅ − =4 6 24 
− − = −5 5 10 ( ) ( )− ⋅ − =5 5 25 
− − = −1 12 13 ( ) ( )− ⋅ − =1 12 12 
− − = −3 4 7 ( ) ( )− ⋅ − =3 4 12 
− − = −2 6 8 ( ) ( )− ⋅ − =2 6 12 
 
Since none of the numbers that add to the sum of −10 , when multiplied, has a product of 12 
and none of the factors of 12 , when added, has a sum of −10 . Therefore, we conclude that 
the trinomial x x2 10 12− + is not factorable using integers, or it is prime. 
Note: A prime polynomial is one that is not factorable using integers. For example, 4 6 92x x+ + , 
2 72y y+ + , 6 2 52w w+ − , x y2 27+ , y y2 6 2− + , and 4 92x + are prime polynomials. 
Example 1.3-29: Factor x x2 4 5+ − . 
Solution: 
To factor the above trinomial we need to obtain two numbers whose sum is 4 and whose 
product is −5 . Let’s construct a table as follows: 
 
Mastering Algebra - Advanced Level 1.3c Factoring Polynomials Using the Trial and Error Method 
Hamilton Education Guides 55 
Sum Product 
5 1 4− = ( )5 1 5⋅ − = − 
 
In this case, at first trial we obtained the sum and the product of the two numbers that we 
need. Thus, x x2 4 5+ − = ( )( )x x+ −5 1 
Check: ( ) ( )x x+ −5 1 = ( )x x x x⋅ − ⋅ + ⋅ + ⋅ −1 5 5 1 = x x x2 5 5− + − = ( )x x2 5 1 5+ − − = x x2 4 5+ − 
Example 1.3-30: Factor x x2 19 66− − . 
Solution: 
To factor the above trinomial we need to obtain two numbers whose sum is −19 and whose 
product is −66 . Let’s construct a table as follows: 
Sum Product 
10 29 19− = − ( )10 29 290⋅ − = − 
9 28 19− = − ( )9 28 252⋅ − = − 
8 27 19− = − ( )8 27 216⋅ − = − 
7 26 19− = − ( )7 26 182⋅ − = − 
6 25 19− = − ( )6 25 150⋅ − = − 
5 24 19− = − ( )5 24 120⋅ − = − 
4 23 19− = − ( )4 23 92⋅ − = − 
3 22 19− = − ( )3 22 66⋅ − = − 
The last line contains the sum and the product of the two numbers that we need. Thus, 
x x2 19 66− − = ( )( )x x+ −3 22 
Check: ( ) ( )x x+ −3 22 = ( ) ( )x x x x⋅ + − ⋅ + ⋅ + ⋅ −22 3 3 22 = x x x2 22 3 66− + − = ( )x x2 22 3 66+ − + − 
 = x x2 19 66− − 
Practice Problems - Factoring Trinomials of the Form ax bx c2 + + where a = 1 
Section 1.3c Case I Practice Problems - Factor the following trinomials using the Trial and 
Error method: 
1. x x2 2 15− − 
 
2. y y2 9 8− + 3. t t2 2 15+ − 
4. y y2 2 11− + 
 
5. x x2 10 21+ + 6. u u2 4 32+ − 
 
Mastering Algebra - Advanced Level 1.3c Factoring Polynomials Using the Trial and Error Method 
Hamilton Education Guides 56 
Case II Factoring Trinomials of the Form ax bx c2 + + where a 1 
To express a trinomial of the form ax bx c2 + + , where a 1 , in its factored form ( ) ( )lx m kx n+ + , 
let us consider the product ( ) ( )lx m kx n+ + and use the FOIL method to see how each term of the 
resulting trinomial is formed ,e.g., 
( ) ( )lx m kx n+ + = ( ) ( ) ( )k l x x l n x k m x m n⋅ ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ + ⋅ = ( ) ( )kl x km x mn2 + + +ln 
Note that the product of the coefficient of the x2 term and the constant term is kl mn⋅ . In 
addition, the product of the coefficients of x is also kl mn⋅ . We use this concept in order to 
express trinomials of the form ax bx c2 + + , where a 1 , in their equivalent factored form. The 
following show the steps in factoring this class of trinomials: 
Step 1 Obtain two numbers m and n whose sum equals to b and whose product equals to 
a c⋅ . 
Step 2 Rewrite the middle term of the trinomial as the sum of the two numbers found in 
Step 1. 
Step 3 Write the trinomial in its factored form by grouping the first two terms and the last 
two terms (see Section 1.3b). Check the answer by using the FOIL method. 
Examples with Steps 
The following examples further illustrate how to factor trinomials of the form ax bx c2 + + , where 
a 1 , using the Trial and Error method: 
Example 1.3-31 
 Factor 6 23 202x x+ + . 
Solution: 
 Step 1 Obtain two numbers whose sum is 23 and whose product is 6 20 120⋅ = . 
 Let’s construct a table as follows: 
Sum Product 
12 11 23+ = 12 11 132⋅ = 
13 10 23+ = 13 10 130⋅ = 
14 9 23+ = 14 9 126⋅ = 
15 8 23+ = 15 8 120⋅ = 
 
 The last line contains the sum and the product of the two numbers that we 
 need. Therefore, 
 Step 2 6 23 202x x+ + = ( )6 15 8 202x x+ + + = 6 15 8 202x x x+ + + = ( ) ( )3 2 5 4 2 5x x x+ + + 
 
 Step 3 ( ) ( )3 2 5 4 2 5x x x+ + + = ( )( )2 5 3 4x x+ + 
 Check: ( )( )2 5 3 4x x+ + = ( ) ( ) ( ) ( )2 3 2 4 5 3 5 4⋅ ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ + ⋅x x x x = 6 8 15 202x x x+ + + 
 = ( )6 8 15 202x x+ + + = 6 23 202x x+ + 
Mastering Algebra - Advanced Level 1.3c Factoring Polynomials Using the Trial and Error Method 
Hamilton Education Guides 57 
Example 1.3-32 
 Factor 10 9 912x x− − . 
Solution: 
 Step 1 Obtain two numbers whose sum is −9 and whose product is ( )10 91 910⋅ − = − . 
 Let’s construct a table as follows: 
Sum Product 
20 29 9− = − ( )20 29 580⋅ − = − 
21 30 9− = − ( )21 30 630⋅ − = − 
22 31 9− = − ( )22 31 682⋅ − = − 
23 32 9− = − ( )23 32 736⋅ − = − 
24 33 9− = − ( )24 33 792⋅ − = − 
25 34 9− = − ( )25 34 850⋅ − = − 
26 35 9− = − ( )26 35 910⋅ − = − 
 
 The last line contains the sum and the product of the two numbers that we 
 need. Therefore, 
 Step 2 10 9 912x x− − = ( )10 26 35 912x x+ − − = 10 26 35 912x x x+ − − = ( ) ( )2 5 13 7 5 13x x x+ − + 
 
 Step 3 ( ) ( )2 5 13 7 5 13x x x+ − + = ( )( )5 13 2 7x x+ − 
 Check: ( )( )5 13 2 7x x+ − = ( ) ( ) ( ) ( )5 2 5 7 13 2 13 7⋅ ⋅ ⋅ + ⋅ − ⋅ + ⋅ ⋅ + ⋅ −x x x x = 10 35 26 912x x x− + − 
 =( )10 35 26 912x x+ − + − = 10 9 912x x− − 
Additional Examples - Factoring Trinomials of the Form ax bx c2 + + where a 1 
The following examples further illustrate how to factor trinomials using the Trial and Error 
method: 
Example 1.3-33: Factor 6 16 102x x+ + . 
Solution: 
To factor the above trinomial we need to obtain two numbers whose sum is 16 and whose 
product is 6 10 60⋅ = . Let’s construct a table as follows: 
Sum Product 
8 8 16+ = 8 8 64⋅ = 
9 7 16+ = 9 7 63⋅ = 
10 6 16+ = 10 6 60⋅ = 
 
The last line contains the sum and the product of the two numbers that we need. Therefore, 
6 16 102x x+ + = ( )6 10 6 102x x+ + + = 6 10 6 102x x x+ + + = ( ) ( )2 3 5 2 3 5x x x+ + + = ( )( )3 5 2 2x x+ + 
Check: ( )( )3 5 2 2x x+ + = ( ) ( ) ( )3 2 3 2 2 5 5 2⋅ ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ + ⋅x x x x = 6 6 10 102x x x+ + + 
 = ( )6 6 10 102x x+ + + = 6 16 102x x+ + 
 
Mastering Algebra - Advanced Level 1.3c Factoring Polynomials Using the Trial and Error Method 
Hamilton Education Guides 58 
Example 1.3-34: Factor 5 8 32x x+ + . 
Solution: 
To factor the above trinomial we need to obtain two numbers whose sum is 8 and whose 
product is 5 3 15⋅ = . Let’s construct a table as follows: 
Sum Product 
1 7 8+ = 1 7 7⋅ = 
2 6 8+ = 2 6 12⋅ = 
3 5 8+ = 3 5 15⋅ = 
4 4 8+ = 4 4 16⋅ = 
 
The third line contains the sum and the product of the two numbers that we need. Therefore, 
5 8 32x x+ + = ( )5 3 5 32x x+ + + = 5 3 5 32x x x+ + + = ( ) ( )x x x5 3 5 3+ + + = ( )( )5 3 1x x+ + 
Check: ( )( )5 3 1x x+ + = ( ) ( ) ( )5 1 5 1 3 1 3 1⋅ ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ + ⋅x x x x = 5 5 3 32x x x+ + + = ( )5 5 3 32x x+ + + 
 = 5 8 32x x+ + 
Example 1.3-35: Factor 6 19 102x x+ + . 
Solution: 
To factor the above trinomial we need to obtain two numbers whose sum is 19 and whose 
product is 6 10 60⋅ = . Let’s construct a table as follows: 
Sum Product 
1 18 19+ = 1 18 18⋅ = 
2 17 19+ = 2 17 34⋅ = 
3 16 19+ = 3 16 48⋅ = 
4 15 19+ = 4 15 60⋅ = 
5 14 19+ = 5 14 70⋅ = 
 
The fourth line contains the sum and the product of the two numbers that we need. Thus, 
6 19 102x x+ + = ( )6 4 15 102x x+ + + = 6 4 15 102x x x+ + + = ( ) ( )2 3 2 5 3 2x x x+ + + = ( )( )2 5 3 2x x+ + 
Check: ( )( )2 5 3 2x x+ + = ( ) ( ) ( )2 3 2 2 5 3 5 2⋅ ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ + ⋅x x x x = 6 4 15 102x x x+ + + 
 = ( )6 4 15 102x x+ + + = 6 19 102x x+ + 
Example 1.3-36: Factor 2 13 152w w− + . 
Solution: 
To factor the above trinomial we need to obtain two numbers whose sum is −13 and whose 
product is 2 15 30⋅ = . Let’s construct a table as follows: 
Sum Product 
− − = −1 12 13 ( ) ( )− ⋅ − =1 12 12 
− − = −2 11 13 ( ) ( )− ⋅ − =2 11 22 
− − = −3 10 13 ( ) ( )− ⋅ − =3 10 30 
− − = −4 9 13 ( ) ( )− ⋅ − =4 9 36 
 
The third line contains the sum and the product of the two numbers that we need. Thus, 
2 13 152w w− + = ( )2 3 10 152w w+ − − + = 2 3 10 152w w w− − + = ( ) ( )w w w2 3 5 2 3− − − = ( )( )2 3 5w w− − 
Mastering Algebra - Advanced Level 1.3c Factoring Polynomials Using the Trial and Error Method 
Hamilton Education Guides 59 
Check: ( )( )2 3 5w w− − = ( ) ( ) ( ) ( )2 1 2 5 3 1 3 5⋅ ⋅ ⋅ + ⋅ − ⋅ + − ⋅ ⋅ + − ⋅ −w w w w = 2 10 3 152w w w− − + 
 = ( )2 10 3 152w w+ − − + = 2 13 152w w− + 
Example 1.3-37: Factor 5 16 32y y− + . 
Solution: 
To factor the above trinomial we need to obtain two numbers whose sum is −16 and whose 
product is 5 3 15⋅ = . Let’s construct a table as follows: 
Sum Product 
− − = −5 11 16 ( ) ( )− ⋅ − =5 11 55 
− − = −4 12 16 ( ) ( )− ⋅ − =4 12 48 
− − = −3 13 16 ( ) ( )− ⋅ − =3 13 39 
− − = −2 14 16 ( ) ( )− ⋅ − =2 14 28 
− − = −1 15 16 ( ) ( )− ⋅ − =1 15 15 
 
The last line contains the sum and the product of the two numbers that we need. Thus, 
5 16 32y y− + = ( )5 1 15 32y y+ − − + = 5 15 32y y y− − + = ( ) ( )y y y5 1 3 5 1− − − = ( )( )5 1 3y y− − 
Check: ( )( )5 1 3y y− − = ( ) ( ) ( )5 5 3 1 3⋅ ⋅ + ⋅ − ⋅ − + − ⋅ −y y y y = 5 15 32y y y− − + = ( )5 15 1 32y y+ − − + 
 = 5 16 32y y− + 
Practice Problems - Factoring Trinomials of the Form ax bx c2 + + where a 1 
Section 1.3c Case II Practice Problems - Factor the following trinomials using the Trial and 
Error method: 
1. 10 11 352x x+ − 
 
2. 6 122x x− − 3. − + +7 46 212x x 
4. 6 11 32 2x xy y− + y is variable 
 
5. 6 402x x+ − 6. 2 3 272x x+ − 
 
Mastering Algebra - Advanced Level 1.3d Other Factoring Methods for Polynomials 
Hamilton Education Guides 60 
1.3d Other Factoring Methods for Polynomials 
The key to successful factorization of polynomials is recognition and use of the right factoring 
method. In this section we will review how to factor binomials of the form a b2 2− (Case I), 
a b3 3± (case II), including Perfect Square Trinomials (Case III) by using formulas that reduce the 
binomials to lower product terms. 
Case I Factoring Polynomials Using the Difference of Two Squares Method 
Binomials of the form a b2 2− are factored to product of two first degree binomials using the 
following factorization method: 
( )( )a b a b a b2 2− = − + 
Note that a b2 2+ is a prime polynomial and can not be factored. The difference of two square 
terms can be factored using the following steps: 
Step 1 Factor the common terms and write the binomial in the standard form of a b2 2− . 
Step 2 Write the binomial in its equivalent factorable form. Check the answer using the 
FOIL method. 
Examples with Steps 
The following examples show the steps as to how binomials of the form a b2 2− are factored: 
Example 1.3-38 Factor 5 31254k − completely. 
Solution: 
 Step 1 5 31254k − = ( )5 6254k − = 5 252 22k −  
 
 Step 2 5 252 2
2
k −


 = ( )( )5 25 252 2k k− + = ( )( )5 5 252 2 2k k− + = ( )( )( )5 5 5 252k k k− + + 
 Check: ( )( )( )5 5 5 252k k k− + + = ( )( )5 5 5 5 5 252k k k k k⋅ + ⋅ − ⋅ − ⋅ + = ( )( )5 5 5 25 252 2k k k k+ − − + 
 = ( )( )5 25 252 2k k− + = ( )5 25 25 25 252 2 2 2k k k k⋅ + ⋅ − ⋅ − ⋅ = ( )5 25 25 6254 2 2k k k+ / / − / / −/ / 
 = ( )5 6254k − = 5 31254k − 
Example 1.3-39 Factor 81 4 4m n− completely. 
Solution: 
 Step 1 81 4 4m n− = ( )9 2 2 22m n− 
 
 Step 2 ( )9 2 2 22m n− = ( )( )9 92 2 2 2m n m n− + = ( )( )3 92 2 2 2 2m n m n− + 
 
 = ( )[ ]( )3 92 2 2 2m n m n− + = ( )( )( )3 3 9 2 2m n m n m n− + + 
Mastering Algebra - Advanced Level 1.3d Other Factoring Methods for Polynomials 
Hamilton Education Guides 61 
 Check: ( )( )( )3 3 9 2 2m n m n m n− + + = ( )( )3 3 3 3 9 2 2m m m n m n n n m n⋅ + ⋅ − ⋅ − ⋅ + 
 = ( )( )9 3 3 92 2 2 2m mn mn n m n+ − − + = ( )( )9 92 2 2 2m n m n− + 
 = 9 9 9 92 2 2 2 2 2 2 2m m m n n m n n⋅ + ⋅ − ⋅ − ⋅ = 81 9 94 2 2 2 2 4m m n m n n+ / − / −/ / / / = 81 4 4m n− 
 
Additional Examples - Factoring Polynomials Using the Difference of Two Squares Method 
The following examples further illustrate how to factor binomials of the form a b2 2− : 
Example 1.3-40: 
4 646 2x x− = ( )4 162 4x x − = 4 42 2 22x x −  = ( )( )4 4 42 2 2x x x− + = ( )( )4 2 42 2 2 2x x x− + 
 
= ( )( )( )4 2 2 42 2x x x x− + + 
Check: ( )( )( )4 2 2 42 2x x x x− + + = ( )( )4 2 2 2 2 42 2x x x x x x⋅ + ⋅ − ⋅ − ⋅ + = ( )( )4 2 2 4 42 2 2x x x x x+ − − + 
 = ( )( )4 4 42 2 2x x x− + = ( )4 4 4 4 42 2 2 2 2x x x x x⋅ + ⋅ − ⋅ − ⋅ = ( )4 4 4 162 4 2 2x x x x+ / − / −/ / 
 = ( )4 162 4x x − = 4 4 162 4 2x x x⋅ − ⋅ = 4 646 2x x− 
Example 1.3-41: 
w8 256− = w4 2
2
16− = ( )( )w w4 416 16− + = ( )w w2 2 42 4 16−  + = ( )( )( )w w w2 2 44 4 16− + + 
 
= ( )( )( )w w w2 2 2 42 4 16− + + = ( )( )( )( )w w w w− + + +2 2 4 162 4 
Check: ( )( )( )( )w w w w− + + +2 2 4 162 4 = ( )( )( )w w w w w w⋅ + ⋅ − ⋅ − ⋅ + +2 2 2 2 4 162 4 
 = ( )( )( )w w w w w2 2 42 2 4 4 16+ − − + + = ( )( )( )w w w2 2 44 4 16− + + 
 = ()( )w w w w w2 2 2 2 44 4 4 4 16⋅ + ⋅ − ⋅ − ⋅ + = ( )( )w w w w4 2 2 44 4 16 16+ − − + 
 = ( )( )w w4 416 16− + = w w w w4 4 4 416 16 16 16⋅ + ⋅ − ⋅ − ⋅ = w w w8 4 416 16 256+ / / − / / −/ / = w8 256− 
Example 1.3-42: 
8 2004 2d d− = ( )8 252 2d d − = ( )8 52 2 2d d − = ( )( )8 5 52d d d− + 
Check: ( )( )8 5 52d d d− + = ( )8 5 5 5 52d d d d d⋅ + ⋅ − ⋅ − ⋅ = ( )8 5 5 252 2d d d d+ / / − / / − = ( )8 252 2d d − 
 = 8 8 252 2 2d d d⋅ − ⋅ = 8 2004 2d d− 
Example 1.3-43: 
( )x y2 25− + = ( )[ ] ( )[ ]x y x y− + + +5 5 = ( ) ( )x y x y− − + +5 5 
Check: ( ) ( )x y x y− − + +5 5 = x x x y x x y y y y x y⋅ + ⋅ + ⋅ − ⋅ − ⋅ − ⋅ − ⋅ − ⋅ − ⋅5 5 5 5 5 5 
 = x xy x xy y y x y2 25 5 5 5 25+ / / + / / − / / − − − / / − − = x y y y2 2 5 5 25− − − − = x y y2 2 10 25− − − 
Mastering Algebra - Advanced Level 1.3d Other Factoring Methods for Polynomials 
Hamilton Education Guides 62 
 = ( )x y y2 2 10 25− + + = ( )x y2 25− + 
Example 1.3-44: 
( )u v+ −3 2 2 = ( )[ ] ( )[ ]u v u v+ − + +3 3 = ( ) ( )u v u v+ − + +3 3 = ( ) ( )u v u v− + + +3 3 
Check: ( ) ( )u v u v− + + +3 3 = u u u v u u v v v v u v⋅ + ⋅ + ⋅ − ⋅ − ⋅ − ⋅ + ⋅ + ⋅ + ⋅3 3 3 3 3 3 
 = u uv u uv v v u v2 23 3 3 3 9+ / / + − / / − − / / + + / / + = u v u2 2 6 9− + + = ( )u u v2 26 9+ + − = ( )u v+ −3 2 2 
Example 1.3-45: 
x y y2 2 2 1− − − = ( )x y y2 2 2 1− + + = ( )x y2 21− + = ( )[ ] ( )[ ]x y x y− + + +1 1 = ( )( )x y x y− − + +1 1 
Check: ( )( )x y x y− − + +1 1 = x x x y x x y y y y x y⋅ + ⋅ + ⋅ − ⋅ − ⋅ − ⋅ − ⋅ − ⋅ − ⋅1 1 1 1 1 1 
 = x xy x xy y y x y2 2 1+ / / + / − / / − − − / − − = x y y y2 2 1− − − − = x y y2 2 2 1− − − 
Example 1.3-46: 
( ) ( )x y+ − +2 42 2 = ( ) ( )[ ] ( ) ( )[ ]x y x y+ − + + + +2 4 2 4 = ( )( )x y x y+ − − + + +2 4 2 4 = ( )( )x y x y− − + +2 6 
Check: ( )( )x y x y− − + +2 6 = x x x y x x y y y y x y⋅ + ⋅ + ⋅ − ⋅ − ⋅ − ⋅ − ⋅ − ⋅ − ⋅6 6 2 2 2 6 
 = x xy x xy y y x y2 26 6 2 2 12+ / / + − / / − − − − − = x x x y y y2 26 2 6 2 12+ − − − − − 
 = x x y y2 24 8 12+ − − − = ( )x x y y2 24 8 16 4+ − − + − + = x x y y2 24 4 8 16+ + − − − 
 = ( ) ( )x x y y2 24 4 8 16+ + − + + = ( ) ( )x y+ − +2 42 2 
Practice Problems - Factoring Polynomials Using the Difference of Two Squares Method 
Section 1.3d Case I Practice Problems - Use the Difference of Two Squares method to factor 
the following polynomials: 
1. x x3 16− = 
 
2. ( ) ( )x y+ − +1 32 2 = 3. t t5 81− = 
4. ( )x x y2 210 25+ + − = 
 
5. c c4 29− = 6. p q q2 2 4 4− − − = 
 
Mastering Algebra - Advanced Level 1.3d Other Factoring Methods for Polynomials 
Hamilton Education Guides 63 
Case II Factoring Polynomials Using the Sum and Difference of Two Cubes Method 
To factor binomials of the form a b3 3+ or a b3 3− we use the following formulas: 
( )( )a b a b a ab b3 3 2 2+ = + − + 
( )( )a b a b a ab b3 3 2 2− = − + + 
Students are encouraged to memorize these two formulas in order to successfully factor this class 
of polynomials. The sum and difference of two cubed binomial terms can be factored using the 
following steps: 
Step 1 Write the binomial in the standard form of a b3 3+ or a b3 3− . 
Step 2 Write the binomial in its equivalent factorable form. Check the answer by 
multiplication. 
Examples with Steps 
The following examples show the steps as to how binomials of the form a b3 3+ and a b3 3− are 
factored: 
Example 1.3-47 Factor 3 817 4x x+ . 
Solution: 
 Step 1 3 817 4x x+ = ( )3 274 3x x + = ( )3 34 3 3x x + 
 
 Step 2 ( )3 34 3 3x x + = ( )( )[ ]3 3 3 34 2 2x x x x+ − ⋅ + = ( )( )[ ]3 3 3 94 2x x x x+ − + 
 Check: ( )( )[ ]3 3 3 94 2x x x x+ − + = ( )3 3 9 3 3 3 3 94 2 2x x x x x x x x⋅ − ⋅ + ⋅ + ⋅ − ⋅ + ⋅ 
 = ( )3 3 9 3 9 274 3 2 2x x x x x x− / + / / + / − / / +/ / = ( )3 274 3x x + = 3 27 34 3 4x x x⋅ + ⋅ = 3 814 3 4x x+ + 
 = 3 817 4x x+ 
Example 1.3-48 Factor 2 2503a − . 
Solution: 
 Step 1 2 2503a − = ( )2 1253a − = ( )2 53 3a − 
 
 Step 2 ( )2 53 3a − = ( )( )2 5 5 52 2a a a− + ⋅ + = ( )( )2 5 5 252a a a− + + 
 Check: ( )( )2 5 5 252a a a− + + = ( )2 5 25 5 5 5 5 252 2a a a a a a a⋅ + ⋅ + ⋅ − ⋅ − ⋅ − ⋅ 
 = ( )2 5 25 5 25 1253 2 2a a a a a+ / + / / / − / − / / / −/ / = ( )2 1253a − = 2 2503a − 
Additional Examples - Factoring Polynomials Using the Sum and Difference of Two Cubes Method 
The following examples further illustrate how to factor binomials of the form a b3 3± using the sum 
and difference of two cubes method: 
Example 1.3-49: 
x3 1+ = x3 31+ = ( )( )x x x+ − ⋅ +1 1 12 2 = ( )( )x x x+ − +1 12 
Mastering Algebra - Advanced Level 1.3d Other Factoring Methods for Polynomials 
Hamilton Education Guides 64 
Check: ( )( )x x x+ − +1 12 = x x x x x x x⋅ − ⋅ + ⋅ + ⋅ − ⋅ + ⋅2 21 1 1 1 1 = x x x x x3 2 2 1− + / + − / +/ / = x3 1+ 
Example 1.3-50: 
x3 1− = x3 31− = ( )( )x x x− + ⋅ +1 1 12 2 = ( )( )x x x− + +1 12 
Check: ( )( )x x x− + +1 12 = x x x x x x x⋅ + ⋅ + ⋅ − ⋅ − ⋅ − ⋅2 21 1 1 1 1 = x x x x x3 2 2 1+ + / − − / −/ / = x3 1− 
Example 1.3-51: 
x y3 3 27− = x y3 3 33− = ( )xy 3 33− = ( ) ( )[ ]xy xy xy− + ⋅ +3 3 32 2 = ( ) ( )[ ]xy xy xy− + +3 3 92 
Check: ( ) ( )[ ]xy xy xy− + +3 3 92 = ( ) ( )xy xy xy xy xy xy xy⋅ + ⋅ + ⋅ − ⋅ − ⋅ − ⋅2 23 9 3 3 3 3 9 
 = ( ) ( ) ( )xy xy xy xy xy3 2 23 9 3 9 27+ / + / / / − / − / / / −/ / = ( )xy 3 27− = x y3 3 27− 
Example 1.3-52: 
a a4 − = ( )a a3 1− = ( )a a3 31− = ( )( )[ ]a a a a− + ⋅ +1 1 12 2 = ( )( )[ ]a a a a− + +1 12 
Check: ( )( )[ ]a a a a− + +1 12 = ( )a a a a a a a a⋅ + ⋅ + ⋅ − ⋅ − ⋅ − ⋅2 21 1 1 1 1 = ( )a a a a a a3 2 2 1+ + / − − / −/ / 
 = ( )a a3 1− = a a4 − 
Example 1.3-53: 
r s3 38− = r s3 3 32− = ( )r s3 32− = ( ) ( )[ ]r s r r s s− + ⋅ +2 2 22 2 = ( )( )r s r rs s− + +2 2 42 2 
Check: ( )( )r s r rs s− + +2 2 42 2 = r r r rs r s s r s rs s s⋅ + ⋅ + ⋅ − ⋅ − ⋅ − ⋅2 2 2 22 4 2 2 2 2 4 
 = r r s rs r s rs s3 2 2 2 2 32 4 2 4 8+ / / + / / − / / − / / −/ / / / = r s3 38− 
Example 1.3-54: 
x y3 327+ = x y3 3 33+ = ( )x y3 33+ = ( ) ( )[ ]x y x x y y+ − ⋅ +3 3 32 2 = ( )( )x y x xy y+ − +3 3 92 2 
Check: ( )( )x y x xy y+ − +3 3 92 2 = x x x xy x y y x y xy y y⋅ − ⋅ + ⋅ + ⋅ − ⋅ + ⋅2 2 2 23 9 3 3 3 3 9 
 = x x y xy x y xy y3 2 2 2 2 33 9 3 9 27− / / + / / + / / − / / +/ / / / = x y3 327+ 
Example 1.3-55: 
c d3 327− = c d3 3 33− = ( )c d3 33− = ( ) ( )[ ]c d c c d d− + ⋅ +3 3 32 2 = ( )( )c d c cd d− + +3 3 92 2 
Check: ( )( )c d c cd d− + +3 3 92 2 = c c c cd c d d c d cd d d⋅ + ⋅ + ⋅ − ⋅ − ⋅ − ⋅2 2 2 23 9 3 3 3 3 9 
 = c c d cd c d cd d3 2 2 2 2 33 9 3 9 27+ / / + / / − / / − / / −/ / / / = c d3 327− 
Practice Problems - Factoring Polynomials Using the Sum and Difference of Two Cubes Method 
Section 1.3d Case II Practice Problems - Use the sum and difference of two cubes method to 
factor the following polynomials: 
1. 4 46x + = 
 
2. x y6 6 8+ = 3. ( )x y+ −2 3 3 = 
4. 2 1286r − = 
 
5. ( )x y− +7 3 3 = 6. x y x y6 5 3 2+ = 
 
Mastering Algebra - Advanced Level 1.3d Other Factoring Methods for Polynomials 
Hamilton Education Guides 65 
Case III Factoring Perfect Square Trinomials 
Trinomials of the form a ab b2 22+ + and a ab b2 22− + are called perfect square trinomials. Note 
that these types of polynomials are easy to recognize because their first and last terms are always 
square and their middle term is twice the product of the quantities being squared in the first and 
last terms. Once perfect square trinomials are identified, they can then be represented in their 
equivalent factored form as shown below: 
( )a ab b a b2 2 22+ + = + 
 
( )a ab b a b2 2 22− + = − 
For example, 36 24 42x x− + , 1 8 16 2+ +y y , 25 30 92 2x xy y+ + , and 49 70 252 2m mn n− + are perfect 
square trinomials because: 
1. Their first term is a square, i.e., ( )62x ; ( )1 2 ; ( )5 2x ; ( )7 2m . 
2. Their last term is a square, i. e., ( )−2 2 ; ( )4 2y ; ( )3 2y ; ( )−5 2n , and 
3. Their middle term is twice the product of the quantities being squared in the first and last 
terms, i.e., ( )2 6 2⋅ ⋅ −x ; ( )2 1 4⋅ ⋅ y ; ( )2 5 3⋅ ⋅x y ; ( )2 7 5⋅ ⋅ −m n . 
Therefore, the above examples can be represented in their equivalent factored form as: ( )6 2 2x − ; 
( )1 4 2+ y ; ( )5 3 2x y+ ; and ( )7 5 2m n− , respectively. 
The following show the steps as to how perfect square trinomials are represented in their equivalent 
factored form: 
Step 1 Write the trinomial in descending order. 
Step 2 Check and see if the trinomial match the general forms a ab b2 22+ + or a ab b2 22− + . 
Step 3 Write the trinomial in its equivalent form, i.e., ( )a b+ 2 or ( )a b− 2 . 
Examples with Steps 
The following examples show the steps as to how perfect square trinomials are factored: 
Example 1.3-56 
 Factor 25 16 402y y+ + . 
Solution: 
 Step 1 25 16 402y y+ + = 25 40 162y y+ + 
 
 Step 2 25 40 162y y+ + = 5 40 42 2 2y y+ + = ( ) ( )5 2 5 4 42 2y y+ ⋅ ⋅ + 
 
 Step 3 ( ) ( )5 2 5 4 42 2y y+ ⋅ ⋅ + = ( )5 4 2y + 
Example 1.3-57 
 Factor 16 24 92 2x xy y+ + . 
Solution: 
 Step 1 Not Applicable 
Mastering Algebra - Advanced Level 1.3d Other Factoring Methods for Polynomials 
Hamilton Education Guides 66 
 
 Step 2 16 24 92 2x xy y+ + = 4 24 32 2 2 2x xy y+ + = ( ) ( ) ( )4 2 4 3 32 2x x y y+ ⋅ ⋅ + 
 
 Step 3 ( ) ( ) ( )4 2 4 3 32 2x x y y+ ⋅ ⋅ + = ( )4 3 2x y+ 
Additional Examples - Factoring Perfect Square Trinomials 
The following examples further illustrate how to factor perfect square trinomials: 
Example 1.3-58: 
x x2 4 4+ + = x x2 24 2+ + = ( )x x2 22 2 2+ ⋅ ⋅ + = ( )x + 2 2 
Example 1.3-59: 
9 6 2− +y y = y y2 6 9− + = ( )y y2 22 3 3− ⋅ ⋅ + = ( )y − 3 2 
Example 1.3-60: 
12 9 4 2t t+ + = 4 12 92t t+ + = 2 12 32 2 2t t+ + = ( ) ( )2 2 2 3 32 2t t+ ⋅ ⋅ + = ( )2 3 2t + 
Example 1.3-61: 
16 40 252x x− + = 4 40 52 2 2x x− + = ( ) ( )4 2 4 5 52 2x x− ⋅ ⋅ + = ( )4 5 2x − 
Example 1.3-62: 
10 1 25 2x x+ + = 25 10 12x x+ + = 5 10 12 2 2x x+ + = ( ) ( )5 2 5 1 12 2x x+ ⋅ ⋅ + = ( )5 1 2x + 
Example 1.3-63: 
4 9 122+ −t t = 9 12 42t t− + = 3 12 22 2 2t t− + = ( ) ( )3 2 3 2 22 2t t− ⋅ ⋅ + = ( )3 2 2t − 
Example 1.3-64: 
9 30 252 2p pq q− + = 3 30 52 2 2 2p pq q− + = ( ) ( ) ( )3 2 3 5 52 2p p q q− ⋅ ⋅ + = ( )3 5 2p q− 
Example 1.3-65: 
121 88 164 2 2 4u u v v− + = 11 88 42 2 2 2 2 2
2 2
u u v v− + = ( ) ( ) ( )11 2 11 4 42 2 2 2 2 2u u v v− ⋅ ⋅ + = ( )11 42 2 2u v− 
Practice Problems - Factoring Perfect Square Trinomials 
Section 1.3d Case III Practice Problems - Factor the following trinomials: 
1. x x2 18 81+ + = 
 
2. 9 64 482+ −p p = 3. 9 25 302w w+ + = 
4. 25 102+ −k k = 
 
5. 49 84 362x x− + = 6. 1 16 64 2+ +z z = 
 
Mastering Algebra - Advanced Level 1.4a Quadratic Equations and the Quadratic Formula 
Hamilton Education Guides 67 
1.4 Quadratic Equations and Factoring 
 
In this section the different methods for factoring quadratic equations are reviewed. The 
Quadratic Formula and its use for solving quadratic equations is addressed in Section 1.4a. 
Solving quadratic equations using the Quadratic Formula, the Square Root Property, and 
Completing the Square method are discussed in Sections 1.4b, 1.4c, and 1.4d, respectively. 
Selection of the best factoring method for solving polynomials or quadratic equations is 
discussed in Section 1.4e. 
1.4a Quadratic Equations and the Quadratic Formula 
A quadratic equation is an equation in which the highest power of the variable is 2 . For 
example, 3 16 5 02x x− + = , x2 16= , w w2 9 0+ = , x x2 4 3 0− + = , x x2 11 24= − − , and y2 4 0− = are all 
examples of quadratic equations. Note that any equation that can be written in the form of 
ax bx c2 0+ + = , where a , b , and c are real numbers and a ≠ 0 , is called a quadratic equation. 
A quadratic equation represented in the form of ax bx c2 0+ + = is said to be in its standard form. 
In the following sections we will review how to solve and represent the solutions to quadratic 
equations in factored form. However, in order to solve any quadratic equation we first need to 
become familiar with the quadratic formula. 
The Quadratic Formula 
To derive the quadratic formula we start with the standard quadratic equation ax bx c2 0+ + = , 
where a , b , and c are real numbers and use the method of completing the square to solve the 
equation as follows: 
Step 1 Add −c to both sides of the equation. 
 ax bx c c c2 + + − = − ; ax bx c2 + = − 
Step 2 Divide both sides of the equation by a . 
 ax
a
bx
a
c
a
2
+ = − ; x bx
a
c
a
2 + = − 
Step 3 Divide b
a
, the coefficient of x , by 2 and square the term to obtain b
a2
2




. Add 
b
a2
2




 to both sides of the equation. 
 x b
a
x b
a
c
a
b
a
2
2 2
2 2
+ + 



= − + 



 
Step 4 Write the left hand side of the equation , which is a perfect square trinomial, in its 
equivalent square form. 
 x b
a
c
a
b
a
+



= − + 


2 2
2 2
 
Step 5 Simplify the right hand side of the equation using the fraction techniques. 
 x b
a
c
a
b
a
+



= − + 


2 2
2 2
 ; x b
a
c
a
b
a
+



= − +
2 4
2 2
2 ; 
( ) ( )
x b
a
a c a b
a a
+



=
⋅ − + ⋅
⋅2
4
4
2 2 2
2 
 ; x b
a
ab a c
a
+



=
−
2
4
4
2 2 2
3 ; 
( )
x b
a
a b ac
a
+



=
−
2
4
4
2 2
3 ; x
b
a
b ac
a
+



=
−
2
4
4
2 2
2 
Step 6 Take the square root of both sides of the equation. 
Mastering Algebra - Advanced Level 1.4a Quadratic Equations and the Quadratic Formula 
Hamilton Education Guides 68 
 x b
a
b ac
a
+



= ±
−
2
4
4
2 2
2 ; x
b
a
b ac
a
+ = ±
−
2
4
2
2
2 2
 ; x b
a
b ac
a
+ = ±
−
2
4
2
2
 
Step 7 Solve for x by adding − b
a2
 to both sides of the equation. 
 x b
a
b
a
b
a
b ac
a
+ − = − ±
−
2 2 2
4
2
2
 ; x b
a
b ac
a
= − ±
−
2
4
2
2
 ; x b b ac
a
=
− ± −2 4
2
 
The equation x b b ac
a
=
− ± −2 4
2
 is referred to as the quadratic formula. Note that the quadratic 
formula has two solutions x b b ac
a
=
− + −2 4
2
 and x b b ac
a
=
− − −2 4
2
. We use these solutions to 
write the quadratic equation ax bx c2 0+ + = in its equivalent factored form, i.e., 
 ax bx c2 0+ + = is factorable to x b b ac
a
x b b ac
a
−
− + −







+
+ −







=
2 24
2
4
2
0 
Let’s check the above factored product using the FOIL method. The result should be equal to 
ax bx c2 0+ + = . 
Check: x b b ac
a
x b b ac
a
−
− + −







+
+ −







=
2 24
2
4
2
0 ; x b b ac
a
x b b ac
a
+
− −







+
+ −







=
2 24
2
4
2
0 
; x x b b ac
a
x b b ac
a
x b b ac
a
b b ac
a
⋅ +
+ −







⋅ +
− −







⋅ +
+ −







⋅
− −







=
2 2 2 24
2
4
2
4
2
4
2
0 
; x b b ac
a
b b ac
a
x
b b ac b b ac
a a
2
2 2
2 2
4
2
4
2
4 4
2 2
0+ + − + − −








+
+ −

 − −




⋅










= 
; x b b ac b b ac
a
x
b b b ac b b ac b ac b ac
a
2
2 2
2 2 2 2 2
2
4 4
2
4 4 4 4
4
0+ + − + − −








+
− − + − − − ⋅ −












= 
; 
( )
x b b
a
x
b b ac
a
2
2 2
22
4
4
0+ +



+
− −









= ; x b
a
x b b ac
a
2
2 2
2
2
2
4
4
0+ 



+
− +




 = ; x
b
a
x ac
a
2
2
4
4
0+ + = 
; x b
a
x c
a
2 0+ + = ; x bx
a
c
a
2
1
0+ + = ; ax bx c
a
2
0+ + = ; ax bx c
a
2 0
1
+ +
= ; ( )ax bx c a2 1 0+ + ⋅ = ⋅ which 
is the same as ax bx c2 0+ + = . 
The quadratic formula is a powerful formula and should be memorized. In the following sections 
we will use this formula to solvedifferent types of quadratic equations. 
Practice Problems - Quadratic Equations and the Quadratic Formula 
Section 1.4a Practice Problems - Given the following quadratic equations identify the coefficients 
a , b , and c . 
1. 3 5 2 2x x= − + 2. 2 52x = 3. 3 5 22w w− = 
4. 15 32= − −y 5. x x2 3 5+ = 
 
6. − + =u u2 2 3 
 
Mastering Algebra - Advanced Level 1.4b Solving Quadratic Equations Using the Quadratic Formula 
Hamilton Education Guides 69 
1.4b Solving Quadratic Equations Using the Quadratic Formula 
As was stated earlier, the quadratic formula can be used to solve any quadratic equation by 
expressing the equation in the standard form of ax bx c2 0+ + = and by substituting the equivalent 
numbers for a , b , and c into the quadratic formula. In this section we will review how to solve 
quadratic equations of the form ax bx c2 0+ + = , where a = 1 (Case I) and where a 〉 1 (Case II), 
using the quadratic formula. 
Case I Solving Quadratic Equations of the Form ax bx c2 0+ + = , where a = 1 , Using the Quadratic Formula 
Quadratic equations of the form ax bx c2 0+ + = , where a = 1, are solved using the following steps: 
Step 1 Write the equation in standard form. 
Step 2 Identify the coefficients a , b , and c . 
Step 3 Substitute the values for a , b , and c into the quadratic equation x b b ac
a
=
− ± −2 4
2
. 
Simplify the equation. 
Step 4 Solve for the values of x . Check the answers by either substituting the x values into 
the original equation or by multiplying the factored product using the FOIL method. 
Step 5 Write the quadratic equation in its factored form. 
 
Examples with Steps 
The following examples show the steps as to how quadratic equations are solved using the 
quadratic formula: 
Example 1.4-1 
 Solve the quadratic equation x x2 5 4+ = − . 
Solution: 
 Step 1 x x2 5 4+ = − ; x x2 5 4 4 4+ + = − + ; x x2 5 4 0+ + = 
 
 Step 2 Let: a = 1 , b = 5 , and c = 4 . Then, 
 
 Step 3 Given: x b b ac
a
=
− ± −2 4
2
 ; x = − ± − × ×
×
5 5 4 1 4
2 1
2
 ; x = − ± −5 25 16
2
 
 
 ; x = − ±5 9
2
 ; x = − ±5 3
2
2
 ; x = − ±5 3
2
 
 
 Step 4 Separate x = − ±5 3
2
 into two equations. 
 I. x = − +5 3
2
 ; x = − /
/
2
2
 ; x = −1
1
 ; x = −1 
 
Mastering Algebra - Advanced Level 1.4b Solving Quadratic Equations Using the Quadratic Formula 
Hamilton Education Guides 70 
 II. x = − −5 3
2
 ; x = − /
/
8
4
2
 ; x = − 4
1
 ; x = −4 
 Check No. 1: I. Let x in x x= − + = −1 5 42 ; ( ) ( )− + × − =−1 5 1 42
?
 ; 1 5 4− =−
?
 ; − = −4 4 
 II. Let x in x x= − + = −4 5 42 ; ( ) ( )− + × − =−4 5 4 42
?
 ; 16 20 4− =−
?
 ; − = −4 4 
 Check No. 2: ( )( )x x x x2 5 4 1 4+ + = + +
?
 ; ( ) ( ) ( ) ( )x x x x x x2 5 4 4 1 1 4+ + = ⋅ + ⋅ + ⋅ + ⋅
?
 
 ; x x x x x2 25 4 4 4+ + = + + +
?
 ; ( )x x x x2 25 4 4 1 4+ + = + + +
?
 ; x x x x2 25 4 5 4+ + = + + 
 Step 5 Therefore, the equation x x2 5 4 0+ + = can be factored to ( )( )x x+ + =1 4 0 . 
Example 1.4-2 
 Solve the quadratic equation x x2 12 35= − − . 
Solution: 
 Step 1 x x2 12 35= − − ; x x x x2 12 12 12 35+ = − + − ; x x2 12 0 35+ = − ; x x2 12 35+ = − 
 
 ; x x2 12 35 35 35+ + = − + ; x x2 12 35 0+ + = 
 
 Step 2 Let: a = 1 , b = 12 , and c = 35 . Then, 
 
 Step 3 Given: x b b ac
a
=
− ± −2 4
2
 ; x = − ± − × ×
×
12 12 4 1 35
2 1
2
 ; x = − ± −12 144 140
2
 
 
 ; x = − ±12 4
2
 ; x = − ±12 2
2
2
 ; x = − ±12 2
2
 
 
 Step 4 Separate x = − ±12 2
2
 into two equations. 
 I. x = − +12 2
2
 ; x = − / /
/
10
5
2
 ; x = − 5
1
 ; x = −5 
 
 II. x = − −12 2
2
 ; x = − / /
/
14
7
2
 ; x = − 7
1
 ; x = −7 
 Check No. 1: I. Let x in x x= − = − −5 12 352 ; ( ) ( )− = − × − −5 12 5 352
?
 ; 25 60 35= −
?
 ; 25 25= 
 II. Let x in x x= − = − −7 12 352 ; ( ) ( )− = − × − −7 12 7 352
?
 ; 49 84 35= −
?
 ; 49 49= 
 Check No. 2: ( )( )x x x x2 12 35 5 7+ + = + +
?
 ; ( ) ( ) ( ) ( )x x x x x x2 12 35 7 5 5 7+ + = ⋅ + ⋅ + ⋅ + ⋅
?
 
 ; x x x x x2 212 35 7 5 35+ + = + + +
?
 ; ( )x x x x2 212 35 7 5 35+ + = + + +
?
 
 ; x x x x2 212 35 12 35+ + = + + 
 Step 5 Therefore, the equation x x2 12 35 0+ + = can be factored to ( )( )x x+ + =5 7 0 . 
Mastering Algebra - Advanced Level 1.4b Solving Quadratic Equations Using the Quadratic Formula 
Hamilton Education Guides 71 
Additional Examples - Solving Quadratic Equations of the Form ax bx c2 0+ + = , where a = 1, Using the Quadratic Formula 
The following examples further illustrate how to solve quadratic equations using the quadratic 
formula: 
Example 1.4-3 
Solve the quadratic equation x x2 16 55= − . 
Solution: 
First, write the equation in standard form, i.e., x x2 16 55 0− + = 
 
Next, let: a = 1 , b = −16 , and c = 55 . Then, 
 
Given: x b b ac
a
=
− ± −2 4
2
 ; 
( ) ( )
x =
− − ± − − × ×
×
16 16 4 1 55
2 1
2
 ; x = ± −16 256 220
2
 ; x = ±16 36
2
 
 
; x = ±16 6
2
2
 ; x = ±16 6
2
 Therefore: 
 
I. x = +16 6
2
 ; x = / /
/
22
11
2
 ; x = 11
1
 ; x = 11 II. x = −16 6
2
 ; x = / /
/
10
5
2
 ; x = 5
1
 ; x = 5 
Check No. 1: I. Let x in x x= = −11 16 552 ; 11 16 11 552 = × −
?
 ; 121 176 55= −
?
 ; 121 121= 
 II. Let x in x x= = −5 16 552 ; 5 16 5 552 = × −
?
 ; 25 80 55= −
?
 ; 25 25= 
Check No. 2: ( )( )x x x x2 16 55 11 5− + = − −
?
 ; ( ) ( ) ( ) ( )x x x x x x2 16 55 5 11 11 5− + = ⋅ + − ⋅ + − ⋅ + − ⋅ −
?
 
 ; x x x x x2 216 55 5 11 55− + = − − +
?
 ; ( )x x x x2 216 55 5 11 55− + = + − − +
?
 
 ; x x x x2 216 55 16 55− + = − + 
Therefore, the equation x x2 16 55 0− + = can be factored to ( )( )x x− − =11 5 0 . 
Example 1.4-4 
Solve the quadratic equation 9 62= − −x x . 
Solution: 
First, write the equation in standard form, i.e., x x2 6 9 0+ + = . 
 
Next, let: a = 1 , b = 6 , and c = 9 . Then, 
 
Given: x b b ac
a
=
− ± −2 4
2
 ; x = − ± − × ×
×
6 6 4 1 9
2 1
2
 ; x = − ± −6 36 36
2
 ; x = − ±6 0
2
 ; x = − ±6 0
2
 
 
; x = − /
/
6
3
2
 ; x = − 3
1
 ; x = −3 
In this case the equation has one repeated solution, i.e., x = −3 and x = −3 . 
Mastering Algebra - Advanced Level 1.4b Solving Quadratic Equations Using the Quadratic Formula 
Hamilton Education Guides 72 
Thus, the solution set is { }− −3 3, . 
Check No. 1: Let x in x x= − + + =3 6 9 02 ; ( )− + × − + =3 6 3 9 02
?
 ; 9 18 9 0− + =
?
 ; 18 18 0− =
?
 ; 0 0= 
Check No. 2: ( )( )x x x x2 6 9 3 3+ + = + +
?
 ; ( ) ( ) ( ) ( )x x x x x x2 6 9 3 3 3 3+ + = ⋅ + ⋅ + ⋅ + ⋅
?
 
 ; x x x x x2 26 9 3 3 9+ + = + + +
?
 ; ( )x x x x2 26 9 3 3 9+ + = + + +
?
 ; x x x x2 26 9 6 9+ + = + + 
Therefore, the equation x x2 6 9 0+ + = can be factored to ( )( )x x+ + =3 3 0 . 
Note that when c = 0 the quadratic equation ax bx c2 0+ + = reduces to ax bx2 0+ = . For cases 
where a = 1, we can solve equations of the form x bx2 0+ = using the quadratic formula in the 
following way: 
Example 1.4-5 
Solve the quadratic equation x x2 5 0+ = . 
Solution: 
The equation is already in standard form. 
 
Let: a = 1 , b = 5 , and c = 0 . Then, 
 
Given: x b b ac
a
=
− ± −2 4
2
 ; x =
− ± − × ×
×
5 5 4 1 0
2 1
2
 ; x =
− ± −5 25 0
2
 ; x =
− ±5 25
2
 ; x =
− ±5 5
2
2
 
 
; x = − ±5 5
2
 Therefore: 
I. x = − +5 5
2
 ; x = 0
2
 ; x = 0 II. x = − −5 5
2
 ; x = − / /
/
10
5
2
 ; x = − 5
1
 ; x = −5 ; x = −5 
and the solution set is { }0 5, − . 
Check No. 1: I. Let x in x x= + =0 5 02 ; 0 5 0 02 + ⋅ =
?
 ; 0 0 0+ =
?
 ; 0 0= 
 II. Let x in x x= − + =5 5 02 ; ( )− + ⋅ − =5 5 5 02
?
 ; 25 25 0− =
?
 ; 0 0= 
Check No. 2: ( )( )x x x x2 5 0 5+ = + +
?
 ; ( ) ( ) ( ) ( )x x x x x x2 5 5 0 0 5+ = ⋅ + ⋅ + ⋅ + ⋅
?
 ; x x x x2 25 5 0 0+ = + + +
?
 
 ; x x x x2 25 5+ = + 
Therefore, the equation x x2 5 0+ = can be factored to ( )( )x x+ + =0 5 0 which is the same as 
( )x x + =5 0 . 
Practice Problems- Solving Quadratic Equations of the Form ax bx c2 + + , where a = 1, Using the Quadratic Formula 
Section 1.4b Case I Practice Problems - Use the quadratic formula to solve the following 
quadratic equations. 
1. x x2 5 6= − − 
 
2. y y2 40 300− = − 3. − = − +x x 2 20 
4. x x2 3 4 0+ + = 5. x x2 80 2 0− − = 6. x x2 4 4 0+ + = 
 
Mastering Algebra – Advanced Level 1.4b Solving Quadratic Equations Using the Quadratic Formula 
Hamilton Education Guides 73 
Case II Solving Quadratic Equations of the Form ax bx c2 0+ + = , where a 1 , Using the Quadratic Formula 
Trinomial equations of the form ax bx c2 0+ + = , where a 1 , are solved using the following steps: 
Step 1 Write the equation in standard form. 
Step 2 Identify the coefficients a , b , and c . 
Step 3 Substitute the values for a , b , and c into the quadratic equation x b b ac
a
=
− ± −2 4
2
. 
Simplify the equation. 
Step 4 Solve for the values of x . Check the answers by either substituting the x values into 
the original equation or by multiplying the factored product using the FOIL method. 
Step 5 Write the quadratic equation in its factored form. 
Examples with Steps 
The following examples show the steps as to how second degree trinomial equations are solved 
using the quadratic formula: 
Example 1.4-6 
 Solve the quadratic equation 2 5 32x x+ = − . 
Solution: 
 Step 1 2 5 32x x+ = − ; 2 5 3 02x x+ + = 
 
 Step 2 Let: a = 2 , b = 5 , and c = 3 . Then, 
 
 Step 3 Given: x b b ac
a
=
− ± −2 4
2
 ; x = − ± − × ×
×
5 5 4 2 3
2 2
2
 ; x = − ± −5 25 24
4
 
 
 ; x = − ±5 1
4
 ; x = − ±5 1
4
 
 
 Step 4 Separate x = − ±5 1
4
 into two equations: 
 I. x = − +5 1
4
 ; x = − /
/
4
4
 ; x = −1
1
 ; x = −1 II. x = − −5 1
4
 ; x = − /
/
6
3
4
2
 ; x = − 32 
 Thus, the solution set is − −




1 3
2
, . 
 Check No. 1: I. Let x in x x= − + = −1 2 5 32 ; ( ) ( )2 1 5 1 32− + × − =−
?
 ; 2 5 3− =−
?
 ; − = −3 3 
 II. Let x in x x= − + = −3
2
2 5 32 ; 2 3
2
5 3
2
3
2
−



+ × −



=−
?
 ; 2 9
4
15
2
3× − =−
?
 
 ; 18
4
15
2
3− =−
?
 ; ( ) ( )2 18 4 15
4 2
3
× − ×
×
=−
?
 ; 36 60
8
3− =−
?
 ; − =−24
8
3
?
 ; − = −3 3 
Mastering Algebra – Advanced Level 1.4b Solving Quadratic Equations Using the Quadratic Formula 
Hamilton Education Guides 74 
 Check No. 2: ( )( )2 5 3 1 2 32x x x x+ + = + +
?
 ; ( ) ( ) ( ) ( )2 5 3 2 3 1 2 1 32x x x x x x+ + = ⋅ + ⋅ + ⋅ + ⋅
?
 
 ; 2 5 3 2 3 2 32 2x x x x x+ + = + + +
?
 ; ( )2 5 3 2 3 2 32 2x x x x+ + = + + +
?
 
 ; 2 5 3 2 5 32 2x x x x+ + = + + 
 Step 5 Therefore, the equation 2 5 3 02x x+ + = can be factored to ( )x x+ +


=1 3
2
0 
 which is the same as ( )( )x x+ + =1 2 3 0 
Example 1.4-7 
 Solve the quadratic equation 15 7 22x x= − + . 
Solution: 
 Step 1 15 7 22x x= − + ; 15 7 7 7 22x x x x+ = − + + ; 15 7 0 22x x+ = + ; 15 7 22x x+ = 
 
 ; 15 7 2 2 22x x+ − = − ; 15 7 2 02x x+ − = 
 
 Step 2 Let: a = 15 , b = 7 , and c = −2 . Then, 
 
 Step 3 Given: x b b ac
a
=
− ± −2 4
2
 ; x = − ± − × × −
×
7 7 4 15 2
2 15
2
 ; x = − ± +7 49 120
30
 
 
 ; x = − ±7 169
30
 ; x = − ±7 13
30
 
 
 Step 4 Separate x = − ±7 13
30
 into two equations: 
 I. x = − +7 13
30
 ; x = /
/ /
6
30
5
 ; x = 15 II. x =
− −7 13
30
 ; x = − / /
/ /
20
2
30
3
 ; x = − 23 
 Thus, the solution set is −




2
3
1
5
, . 
 Check No. 1: I. Let x in x x= = − +1
5
15 7 22 ; 15 1
5
7 1
5
2
2




= − ×



+
?
 ; 15 1
25
7
5
2× =− +
?
 
 ; / /
/ /
=− +
15
3
25
5
7
5
2
1
?
 ; ( ) ( )3
5
7 1 2 5
5 1
=
− × + ×
×
?
 ; 3
5
7 10
5
=
− +? ; 3
5
3
5
= 
 II. Let x in x x= − = − +2
3
15 7 22 ; 15 2
3
7 2
3
2
2
−



= − × −



+
?
 ; 15 4
9
14
3
2× = +
?
 
 ; / /
/
= +
60
20
9
3
14
3
2
1
?
 ; ( ) ( )20
3
14 1 2 3
3 1
=
× + ×
×
?
 ; 20
3
14 6
3
=
+? ; 20
3
20
3
= 
 Check No. 2: ( )( )15 7 2 5 1 3 22x x x x+ − = − +
?
 ; ( ) ( ) ( ) ( )15 7 2 5 3 2 5 1 3 1 22x x x x x x+ − = ⋅ + ⋅ + − ⋅ + − ⋅
?
 
 ; 15 7 2 15 10 3 22 2x x x x x+ − = + − −
?
 ; ( )15 7 2 15 10 3 22 2x x x x+ − = + − −
?
 
 ; 15 7 2 15 7 22 2x x x x+ − = + − 
Mastering Algebra – Advanced Level 1.4b Solving Quadratic Equations Using the Quadratic Formula 
Hamilton Education Guides 75 
 Step 5 Therefore, the equation 15 7 2 02x x+ − = can be factored to x x−



+



=
1
5
2
3
0 
 which is the same as ( )( )5 1 3 2 0x x− + = 
Additional Examples - Solving Quadratic Equations of the Form ax bx c2 0+ + = , where a 1 , Using the Quadratic Formula 
The following examples further illustrate how to solve quadratic equations: 
Example 1.4-8 
Solve the quadratic equation 3 7 6 02x x+ − = . 
Solution: 
The equation is already in standard form. Let: a = 3 , b = 7 , and c = −6 . Then, 
 
Given: x b b ac
a
=
− ± −2 4
2
 ; x = − ± − × × −
×
7 7 4 3 6
2 3
2
 ; x = − ± +7 49 72
6
 ; x = − ±7 121
6
 
 
; x = − ±7 11
6
2
 ; x = − ±7 11
6
 Therefore: 
 
I. x = − +7 11
6
 ; x = /
/
4
2
6
3
 ; x = 23 II. x =
− −7 11
6
 ; x = − / /
/
18
3
6
 ; x = − 3
1
 ; x = −3 
Thus, the solution set is −




3 2
3
, . 
Check No. 1: I. Let x in x x= − + − =3 3 7 6 02 ; ( ) ( )3 3 7 3 6 02⋅ − + ⋅ − − =
?
 ; 3 9 21 6 0⋅ − − =
?
 
 ; 27 21 6 0− − =
?
 ; 27 27 0− =
?
 ; 0 0= 
 II. Let x in x x= + − =2
3
3 7 6 02 ; 3 2
3
7 2
3
6 0
2
⋅ 



+ ⋅ 



− =
?
 ; 3 4
9
14
3
6 0⋅ + − =
?
 
 ; / /
/
+ − =
12
4
9
3
14
3
6 0
?
 ; 4
3
14
3
6 0+ − =
?
 ; 4 14
3
6 0+ − =
?
 ; / /
/
− =
18
6
3
6 0
?
 ; 6
1
6 0− =
?
 ; 6 6 0− =
?
 
 ; 0 0= 
Check No. 2: ( )( )3 7 6 3 3 22x x x x+ − = + −
?
 ; ( ) ( ) ( ) ( )3 7 6 3 2 3 3 3 22x x x x x x+ − = ⋅ + − ⋅ + ⋅ + ⋅ −
?
 
 ; 3 7 6 3 2 9 62 2x x x x x+ − = − + −
?
 ; ( )3 7 6 3 2 9 62 2x x x x+ − = + − + −
?
 
 ; 3 7 6 3 7 62 2x x x x+ − = + − 
Therefore, the equation 3 7 6 02x x+ − = can be factored to ( )x x+ −


=3 2
3
0 which is the same 
as ( )x x+ −


=3
1
2
3
0 ; ( ) ( ) ( )x x+ ⋅ − ⋅
⋅





 =3
3 1 2
1 3
0 ; ( )x x+ −


=3 3 2
3
0 ; x x+



−



=
3
1
3 2
3
0 
; ( ) ( )x x+ ⋅ −
⋅
=
3 3 2
1 3
0 ; ( ) ( )x x+ ⋅ −
⋅
=
3 3 2
1 3
0
1
 ; ( ) ( )[ ]x x+ ⋅ − ⋅ = ⋅3 3 2 1 0 3 ; ( )( )x x+ − =3 3 2 0 
Mastering Algebra - Advanced Level 1.4b Solving Quadratic Equations Using the Quadratic Formula 
Hamilton Education Guides 76 
Example 1.4-9 
Solve the quadratic equation 4 9 62x x+ = − . 
Solution: 
First, write the equation in standard form, i.e., 4 9 6 02x x+ + = . 
 
Next, let: a = 4 , b = 9 , and c = 6 . Then, 
 
Given: x b b ac
a
=
− ± −2 4
2
 ; x = − ± − × ×
×
9 9 4 4 6
2 4
2
 ; x = − ± −9 81 96
8
 ; x = − ± −9 15
8
 
 
Since the number under the radical is negative, therefore the quadratic equation does not have 
any real solutions. We state that the equation is not factorable. 
Note that when c = 0 the quadratic equation ax bx c2 0+ + = reduces to ax bx2 0+ = . For cases 
where a 〉 1 , we can solve equations of the form ax bx2 0+ = using the quadratic formula in the 
following way: 
Example 1.4-10 
Solve the quadratic equation 2 5 02x x+ = . 
Solution: 
First write the equation in standard form, i.e., 2 5 0 02x x+ + = . 
 
Next, let: a = 2 , b = 5 , and c = 0 . Then, 
 
Given: x b b ac
a
=
− ± −2 4
2
 ; x = − ± − × ×
×
5 5 4 2 0
2 2
2
 ; x = − ± −5 25 0
4
 ; x = − ±5 25
4
 
 
; x = − ±5 5
4
2
 ; x = − ±5 5
4
 Therefore: 
 
I. x = − +5 5
4
 ; x = 0
4
 ; x = 0 II. x = − −5 5
4
 ; x = − 5
2
 ; x = −2 5. 
Thus, the solution set is { }0 2 5, .− . 
Check: I. Let x in x x= + =0 2 5 02 ; 2 0 5 0 02⋅ + ⋅ =
?
 ; 0 0 0+ =
?
 ; 0 0= 
 II. Letx in x x= − + =2 5 2 5 02. ; ( )2 2 5 5 2 5 02⋅ − + ⋅ − =. .
?
 ; 2 6 25 12 5 0⋅ − =. .
?
 ; 12 5 12 5. .= 
Therefore, the equation 2 5 02x x+ = can be factored to ( )( )x x+ + =0 2 5 0. which is the same as 
( )x x + =2 5 0. . 
Practice Problems - Solving Quadratic Equations of the Form ax bx c2 0+ + = , where a 1 , Using the Quadratic Formula 
Section 1.4b Case II Practice Problems - Use the quadratic formula to solve the following 
quadratic equations. 
1. 4 6 1 02u u+ + = 
 
2. 4 10 32w w+ = − 3. 6 4 2 02x x+ − = 
4. 15 3 142y y+ = − 
 
5. 2 5 3 02x x− + = 6. 2 02 2x xy y+ − = x is variable 
 
Mastering Algebra - Advanced Level 1.4c Solving Quadratic Equations Using the Square Root Property Method 
Hamilton Education Guides 77 
1.4c Solving Quadratic Equations Using the Square Root Property Method 
Quadratic equations of the form ( )ax b c+ =2 are solved using a method known as the Square 
Root Property method where the square root of both sides of the equation are taken and the terms 
are simplified. Following show the steps as to how quadratic equations are solved using the 
Square Root property method: 
Step 1 Take the square root of the left and the right hand side of the equation. Simplify the 
terms on both sides of the equation. 
Step 2 Solve for the values of x . Check the answers by substituting the x values into the 
original equation. 
Step 3 Write the equation in its factored form. 
Examples with Steps 
The following examples show the steps as to how equations of the form ( )ax b c+ =2 are solved 
using the Square Root Property method: 
Example 1.4-11 
 Solve the quadratic equation ( )x + =4 362 . 
Solution: 
 Step 1 ( )x + =4 362 ; ( )x + = ±4 362 ; ( )x + = ±4 62 2 ; x + = ±4 6 
 
 Step 2 Separate x + = ±4 6 into two equations. 
 I. x + = +4 6 ; x = + −6 4 ; x = 2 
 
 II. x + = −4 6 ; x = − −6 4 ; x = −10 
 Thus, the solution set is { }−10 2, . 
 Check: I. ( )Let x in x= + =2 4 362 ; ( )2 4 362+ =
?
 ; 6 362 =
?
 ; 36 36= 
 II. ( )Let x in x= − + =10 4 362 ; ( )− + =10 4 362
?
 ; ( )− =6 362
?
 ; 36 36= 
 Step 3 Therefore, the equation ( )x + =4 362 can be factored to ( )( )x x− + =2 10 0 . 
Example 1.4-12 
 Solve the quadratic equation ( )x − =2 252 . 
Solution: 
 Step 1 ( )x − =2 252 ; ( )x − = ±2 252 ; ( )x − = ±2 52 2 ; x − = ±2 5 
 
 Step 2 Separate x − = ±2 5 into two equations. 
 I. x − = +2 5 ; x = +5 2 ; x = 7 
 
 II. x − = −2 5 ; x = − +5 2 ; x = −3 
Mastering Algebra - Advanced Level 1.4c Solving Quadratic Equations Using the Square Root Property Method 
Hamilton Education Guides 78 
 Thus, the solution set is { }−3 7, . 
 Check: I. ( )Let x in x= − =7 2 252 ; ( )7 2 252− =
?
 ; 5 252 =
?
 ; 25 25= 
 II. ( )Let x in x= − − =3 2 252 ; ( )− − =3 2 252
?
 ; ( )− =5 252
?
 ; 25 25= 
 Step 3 Therefore, the equation ( )x − =2 252 can be factored to ( )( )x x− + =7 3 0 . 
Additional Examples - Solving Quadratic Equations Using the Square Root Property Method 
The following examples further illustrate how to solve quadratic equations using the Square Root 
Property method: 
Example 1.4-13 
Solve the quadratic equation ( )5 3 152y + = using the Square Root Property method. 
Solution: 
( )5 3 152y + = ; ( )5 3 152y + = ± ; 5 3 15y + = ± Therefore, the two solutions are: 
 
I. 5 3 15y + = + ; 5 15 3y = − ; y = −15 35 II. 5 3 15y + = − ; 5 15 3y = − − ; y = −
+15 3
5 
Thus, the solution set is − + −








15 3
5
15 3
5
, . 
Check: I. ( )Let y in y= − + =15 3
5
5 3 152 ; / ⋅ −
/
+





 =5
15 3
5
3 15
2 ?
 ; ( )15 3 3 152− / + / =? ; ( )15 152 =? 
 ; 15 15= 
 II. ( )Let y in y= − + + =15 3
5
5 3 152 ; / ⋅ − +
/





 +








=5 15 3
5
3 15
2
?
 ; ( )[ ]− − + =15 3 3 152 ? 
 ; ( )− − / + / =15 3 3 152 ? ; ( )− =15 152 ? ; 15 15= 
Therefore, the equation ( )5 3 152y + = can be factored to y y− −





 +
+




 =
15 3
5
15 3
5
0 which is 
the same as ( )( )y y− + =0175 1 375 0. . ; y y2 12 0 24 0+ − =. . ; 25 30 6 02y y+ − = , or ( )5 3 152y + = . 
Example 1.4-14 
Solve the quadratic equation ( )x + =5 492 using the Square Root Property and the Quadratic 
Formula method. 
Solution: 
First Method - The Square Root Property method: 
( )x + =5 492 ; ( )x + = ±5 492 ; ( )x + = ±5 72 2 ; x + = ±5 7 Therefore, the two solutions are: 
 
I. x + = +5 7 ; x = −7 5 ; x = 2 
 
II. x + = −5 7 ; x = − −7 5 ; x = −12 
Mastering Algebra - Advanced Level 1.4c Solving Quadratic Equations Using the Square Root Property Method 
Hamilton Education Guides 79 
Thus, the solution set is { }−12 2, . 
Check: I. ( )Let x in x= + =2 5 492 ; ( )2 5 492+ =
?
 ; 7 492 =
?
 ; 49 49= 
 II. ( )Let x in x= − + =13 5 492 ; ( )− + =12 5 492
?
 ; ( )− =7 492
?
 ; 49 49= 
Therefore, the equation ( )x + =5 492 can be factored to ( )( )x x− + =2 12 0 . 
Second Method - The Quadratic Formula method: 
Given the expression ( )x + =5 492 , expand the left hand side of the equation and write the 
quadratic equation in its standard form, i.e., 
( )x + =5 492 ; x x2 25 10 49+ + = ; ( )x x2 10 25 49 0+ + − = ; x x2 10 24 0+ − = 
 
Let: a = 1 , b = 10 , and c = −24 . Then, 
 
Given: x b b ac
a
=
− ± −2 4
2
 ; 
( )
x =
− ± − × × −
×
10 10 4 1 24
2 1
2
 ; x = − ± +10 100 96
2
 ; x = − ±10 196
2
 
 
; x = − ±10 14
2
2
 ; x = − ±10 14
2
 Therefore, we can separate x into two equations: 
 
I. x = − +10 14
2
 ; x = /
/
4
2
2
 ; x = 2
1
 ; x = 2 II. x = − −10 14
2
 ; x = − / /
/
24
12
2
 ; x = −12
1
 ; x = −12 
Thus, the solution set is { }−12 2, . 
The equation ( )x + =5 492 can be factored to ( )( )x x− + =2 12 0 . 
Note: As you may have already noticed, using the quadratic formula may not be a good 
choice since it requires more work and takes longer to solve. The key to solving quadratic 
equations is selection of a method that is easiest to use. Further discussions on selection of a 
best method is addressed in Section 1.4e. 
Note that when b = 0 the quadratic equation ( )ax b c+ =2 reduces to ( )ax c2 = . The following 
examples show the steps as to how quadratic equations of the form ( )ax c2 = are solved for cases 
where the coefficient of x is equal to or greater than one. 
• For cases where a = 1, we can solve equations of the form x c2 = using the Square Root 
Property method in the following way: 
Example 1.4-15 
Solve x2 16= using the Square Root Property method. 
Solution: 
First - Take the square root of both sides of the equation, i.e., x2 16= ± 
Second - Simplify the terms on both sides to obtain the solutions, i.e., x = ±4 . Therefore, the 
solution set is { }−4 4, and the equation x2 16= can be factored to ( )( )x x− + =4 4 0 . 
Mastering Algebra - Advanced Level 1.4c Solving Quadratic Equations Using the Square Root Property Method 
Hamilton Education Guides 80 
Check: I. Let x in x= − =4 162 ; ( )− =4 162
?
 ; 16 16= 
 II. Let x in x= =4 162 ; 4 162 =
?
 ; 16 16= 
• For cases where a 〉 1 , we can solve equations of the form ( )ax c2 = (which is the same as 
kx c2 = , where k a= 2 ) using the Square Root Property method in the following way: 
Example 1.4-16 
Solve 3 272x = using the Square Root Property method. 
Solution: 
First - Divide both sides of the equation by the coefficient x , i.e., /
/
=
/ /
/
3
3
27
9
3
2x ; x2 9
1
= ; x2 9= 
Second - Take the square root of both sides of the equation, i.e., x2 9= ± 
 
Third - Simplify the terms on both sides to obtain the solutions, i.e., x = ±3 
Therefore, the solution set is { }−3 3, and the equation 3 272x = can be factored to ( )( )x x− + =3 3 0 . 
Check: I. Let x in x= − =3 3 272 ; ( )3 3 272⋅ − =
?; 3 9 27⋅ =
?
 ; 27 27= 
 II. Let x in x= =3 3 272 ; 3 3 272⋅ =
?
 ; 3 9 27⋅ =
?
 ; 27 27= 
Practice Problems - Solving Quadratic Equations Using the Square Root Property Method 
Section 1.4c Practice Problems - Solve the following equations using the Square Root Property 
method: 
1. ( )2 5 362y + = 
 
2. ( )x + =1 72 3. ( )2 3 12x − = 
4. x2 3 0+ = 
 
5. ( )y − =5 52 6. 16 25 02x − = 
 
Mastering Algebra - Advanced Level 1.4d Solving Quadratic Equations Using Completing-the-Square Method 
Hamilton Education Guides 81 
1.4d Solving Quadratic Equations Using Completing-the-Square Method 
One of the methods used in solving quadratic equations is called Completing-the-Square method. 
Note that this method involves construction of perfect square trinomials which was addressed in 
Section 1.3 d, Case III. In this section we will review how to solve quadratic equations of the 
form ax bx c2 0+ + = , where a = 1 (case I) and where a 〉1 (Case II), using Completing-the-Square 
method. 
Case I Solving Quadratic Equations of the Form ax bx c2 0+ + = , where a = 1 , by Completing the Square 
The following show the steps as to how quadratic equations, where the coefficient of the squared 
term is equal to one, are solved using Completing-the-Square method: 
Step 1 Write the equation in the form of x bx c2 + = − . 
Step 2 a. Divide the coefficient of x by 2 , i.e., b
2
. 
 b. Square half the coefficient of x obtained in step 2a, i.e., b
2
2




. 
 c. Add the square of half the coefficient of x to both sides of the equation, i.e., 
 x bx b c b2
2 2
2 2
+ + 



= − + 



. 
 d. Simplify the equation. 
Step 3 Factor the trinomial on the left hand side of the equation as the square of a binomial, 
i.e., x b c b+



= − + 


2 2
2 2
. 
Step 4 Take the square root of both sides of the equation and solve for the x values, i.e., 
x b c b+



= ± − + 


2 2
2 2
 ; x b c b+ = ± − + 


2 2
2
 ; x b c b= − ± − + 


2 2
2
. 
Step 5 Check the answers by substituting the x values into the original equation. 
Step 6 Write the quadratic equation in its factored form. 
Examples with Steps 
The following examples show the steps as to how quadratic equations, where the coefficient of 
the squared term is equal to one, are solved using Completing-the-Square method: 
Example 1.4-17 
 Solve the quadratic equation x x2 6 0+ − = by completing the square. 
Solution: 
 Step 1 x x2 6 0+ − = ; x x2 6 6 6+ − + = + ; x x2 0 6+ + = ; x x2 6+ = 
 
 Step 2 x x2 6+ = ; x x2
2 21
2
6 1
2
+ + 



= + 



 ; x x2 1
4
6 1
4
+ + = + ; x x2 1
4
6
1
1
4
+ + = + 
 
Mastering Algebra - Advanced Level 1.4d Solving Quadratic Equations Using Completing-the-Square Method 
Hamilton Education Guides 82 
 ; ( ) ( )x x2 1
4
6 4 1 1
1 4
+ + =
⋅ + ⋅
⋅
 ; x x2 1
4
24 1
4
+ + =
+ ; x x2 1
4
25
4
+ + = 
 
 Step 3 x x2 1
4
25
4
+ + = ; x +



=
1
2
25
4
2
 
 
 Step 4 x +



=
1
2
25
4
2
 ; x +


= ±
1
2
25
4
2
 ; x + = ±1
2
5
2
2
2 ; x + = ±
1
2
5
2
 therefore: 
 
 I. x + = +1
2
5
2
 ; x = − +1
2
5
2
 ; x = − +1 5
2
 ; x = /
/
4
2
2
 ; x = 2 
 
 II. x + = −1
2
5
2
 ; x = − −1
2
5
2
 ; x = − −1 5
2
 ; x = − /
/
6
3
2
 ; x = −3 
 Thus, the solution set is { }−3 2, . 
 Step 5 Check: Substitute x = 2 and x = −3 in x x2 6 0+ − = 
 I. Let x in x x= + − =2 6 02 ; 2 2 6 02 + − =
?
 ; 4 2 6 0+ − =
?
 ; 6 6 0− =
?
 ; 0 0= 
 II. Let x in x x= − + − =3 6 02 ; ( )− − − =3 3 6 02
?
 ; 9 3 6 0− − =
?
 ; 9 9 0− =
?
 ; 0 0= 
 Step 6 Thus, the equation x x2 6 0+ − = can be factored to ( )( )x x− + =2 3 0 
Example 1.4-18 
 Solve the quadratic equation x x2 2 5 0+ + = by completing the square. 
Solution: 
 Step 1 x x2 2 5 0+ + = ; x x2 2 5 5 5+ + − = − ; x x2 2 0 5+ + = − ; x x2 2 5+ = − 
 
 Step 2 x x2 2 5+ = − ; x x2
2 2
2 2
2
5 2
2
+ +
/
/




= − +
/
/




 ; x x2 2 22 1 5 1+ + = − + 
 
 ; x x2 2 1 5 1+ + = − + ; x x2 2 1 4+ + = − 
 
 Step 3 x x2 2 1 4+ + = − ; ( )x + = −1 42 
 
 Step 4 ( )x + = −1 42 ; ( )x + = ± −1 42 ; x + = ± −1 4 −4 is not a real number. 
 
 Therefore, the equation x x2 2 5 0+ + = does not have any real solutions. 
 
 Step 5 Not Applicable 
 
 Step 6 Not Applicable 
 
Mastering Algebra - Advanced Level 1.4d Solving Quadratic Equations Using Completing-the-Square Method 
Hamilton Education Guides 83 
Additional Examples - Solving Quadratic Equations of the Form ax bx c2 0+ + = , where a = 1, by Completing the Square 
The following examples further illustrate how to solve quadratic equations using Completing the 
Square method: 
Example 1.4-19 
Solve the quadratic equation x x2 3 7 0+ − = using Completing-the-Square method. 
Solution: 
x x2 3 7 0+ − = ; x x2 3 7+ = ; x x2
2 2
3 3
2
7 3
2
+ + 



= + 



 ; x x2 3 9
4
7 9
4
+ + = + ; x +



= +
3
2
7
1
9
4
2
 
 
; ( ) ( )x +



=
⋅ + ⋅
⋅
3
2
7 4 1 9
1 4
2
 ; x +



=
+3
2
28 9
4
2
 ; x +



=
3
2
37
4
2
 ; x +


= ±
3
2
37
4
2
 ; x + = ±3
2
37
2
 
 
therefore: I. x + = +3
2
37
2
 ; x = −37
2
3
2
 ; x = −6 083 3
2
. ; x = 3 083
2
. ; x = 1 541. 
 
 II. x + = −3
2
37
2
 ; x = − −37
2
3
2
 ; x = − −6 083 3
2
. ; x = − 9 083
2
. ; x = −4 541. 
and the solution set is { }−4 541 1 541. , . . 
Check: I. Let x in x x= + − =1541 3 7 02. ; ( ) ( )1541 3 1541 7 02. .
?
+ × − = ; 2 38 4 62 7 0. .
?
+ − = ; 0 0= 
 II. Let x in x x= − + − =4 541 3 7 02. ; ( ) ( )− + × − − =4 541 3 4 541 7 02. .
?
 ; 20 62 1362 7 0. .
?
− − = 
 ; 0 0= 
Therefore, the equation x x2 3 7 0+ − = can be factored to ( )( )x x− + =1 541 4 541 0. . . 
Example 1.4-20 
Solve the quadratic equation x x2 5 6 0+ + = using Completing-the-Square method. 
Solution: 
x x2 5 6 0+ + = ; x x2 5 6+ = − ; x x2
2 2
5 5
2
6 5
2
+ + 



= − + 



 ; x x2 5
25
4
6 25
4
+ + = − + ; x +



= − +
5
2
6
1
25
4
2
 
 
; 
( ) ( )
x +



=
− ⋅ + ⋅
⋅
5
2
6 4 1 25
1 4
2
 ; x +



=
− +5
2
24 25
4
2
 ; x +



=
5
2
1
4
2
 ; x +


= ±
5
2
1
4
2
 ; x + = ±5
2
1
2
 
 
therefore: I. x + = +5
2
1
2
 ; x = −1
2
5
2
 ; x = −1 5
2
 ; x = − /
/
4
2
2
 ; x = − 2
1
 ; x = −2 
 
 II. x + = −5
2
1
2
 ; x = − −1
2
5
2
 ; x = − −1 5
2
 ; x = − /
/
6
3
2
 ; x = − 3
1
 ; x = −3 
and the solution set is { }− −3 2, . 
Mastering Algebra - Advanced Level 1.4d Solving Quadratic Equations Using Completing-the-Square Method 
Hamilton Education Guides 84 
Check: I. Let x in x x= − + + =2 5 6 02 ; ( ) ( )− + × − + =2 5 2 6 02
?
 ; 4 10 6 0− + =
?
 ; 10 10 0− =
?
 ; 0 0= 
 II. Let x in x x= − + + =3 5 6 02 ; ( ) ( )− + × − + =3 5 3 6 02
?
 ; 9 15 6 0− + =
?
 ; 15 15 0− =
?
 ; 0 0= 
Therefore, the equation x x2 5 6 0+ + = can be factored to ( )( )x x+ + =2 3 0 . 
Example 1.4-21 
Solve the quadratic equation y y2 9 11 0− + = using Completing-the-Square method. 
Solution: 
y y2 9 11 0− + = ; y y2 9 11− = − ; y y2
2 2
9 9
2
11 9
2
− + −



= − + −



 ; y y2 9 81
4
11 81
4
− + = − + 
 
; y −



= − +
9
2
11
1
81
4
2
 ; ( ) ( )y −



=
− ⋅ + ⋅
⋅
9
2
11 4 1 81
1 4
2
 ; y −



=
− +9
2
44 81
4
2
 ; y −



=
9
2
37
4
2
 
 
; y −


= ±
9
2
37
4
2
 ; y − = ±9
2
37
2
 therefore: 
 
 I. y − = +9
2
37
2
 ; y = +37
2
9
2
 ; y = +6 083 9
2
. ; y = 15 083
2
. ; y = 7 541. 
 
 II. y − = −9
2
37
2
 ; y = − +37
2
9
2
 ; y = − +6 083 9
2
. ; y = 2 917
2
. ; y = 1 459. 
and the solution set is { }1 459 7 541. , . . 
Check: I. Let y in y y= − + =7 541 9 11 02. ; ( ) ( )7 541 9 7 541 11 02. .
?
+ − × + = ; 56 87 67 87 11 0. .
?
− + =; 67 87 67 87 0. .
?
− = ; 0 0= 
 II. Let y in y y= − + =1459 9 11 02. ; ( ) ( )1459 9 1459 11 02. .
?
+ − × + = ; 213 1313 11 0. .
?
− + = 
 ; 1313 1313 0. .
?
− = ; 0 0= 
Therefore, the equation y y2 9 11 0− + = can be factored to ( )( )y y− − =7 541 1 459 0. . . 
Practice Problems - Solving Quadratic Equations of the Form ax bx c2 0+ + = , where a = 1, by Completing the Square 
Section 1.4d Case I Practice Problems - Solve the following quadratic equations using 
Completing-the-Square method: 
1. x x2 10 2 0+ − = 
 
2. x x2 1 0− − = 3. ( )x x + =2 80 
4. y y2 10 5 0− + = 
 
5. x x2 4 5 0+ − = 6. y y2 4 14+ = 
 
Mastering Algebra - Advanced Level 1.4d Solving Quadratic Equations Using Completing-the-Square Method 
Hamilton Education Guides 85 
Case II Solving Quadratic Equations of the Form ax bx c2 0+ + = , where a 1 , by Completing the Square 
The following show the steps as to how quadratic equations, where the coefficient of the squared 
term is not equal to one, are solved using Completing-the-Square method: 
Step 1 Write the equation in the form of ax bx c2 + = − . 
Step 2 Divide both sides of the equation by a , i.e., /
/
+ = −
ax
a
bx
a
c
a
2
 ; x b
a
x c
a
2 + = − . 
Step 3 a. Divide the coefficient of x by 2 , i.e., 1
2
⋅
b
a
 = 
b
a2
. 
 b. Square half the coefficient of x obtained in step 3a, i.e., b
a2
2




 
 c. Add the square of half the coefficient of x to both sides of the equation, i.e., 
 x b
a
x b
a
c
a
b
a
2
2 2
2 2
+ + 



= − + 



. 
 d. Simplify the equation. 
Step 4 Factor the trinomial on the left hand side of the equation as the square of a binomial, 
i.e., x b
a
c
a
b
a
+



= − + 


2 2
2 2
. 
Step 5 Take the square root of both sides of the equation and solve for the x values, i.e., 
x b
a
c
a
b
a
+



= ± − + 


2 2
2 2
 ; x b
a
c
a
b
a
+ = ± − + 


2 2
2
 ; x b
a
c
a
b
a
= − ± − + 


2 2
2
. 
Step 6 Check the answers by substituting the x values into the original equation. 
Step 7 Write the quadratic equation in its factored form. 
Examples with Steps 
The following examples show the steps as to how quadratic equations, where the coefficient of 
the squared term is not equal to one, are solved using completing-the-square method: 
Example 1.4-22 
 Solve the quadratic equation 2 3 6 02x x+ − = by completing the square. 
Solution: 
 Step 1 2 3 6 02x x+ − = ; 2 3 6 6 62x x+ − + = + ; 2 3 0 62x x+ + = + ; 2 3 62x x+ = 
 
 Step 2 2 3 62x x+ = ; /
/
+ =
/
/
2
2
3
2
6
3
2
2x x ; x x2 3
2
3
1
+ = ; x x2 3
2
3+ = 
 
 Step 3 x x2 3
2
3+ = ; x x2
2 23
2
3
4
3 3
4
+ + 



= + 



 ; x x2 3
2
9
16
3 9
16
+ + = + 
 
 ; x x2 3
2
9
16
3
1
9
16
+ + = + ; x x2 3
2
9
16
48 9
16
+ + =
+ ; x x2 3
2
9
16
57
16
+ + = 
Mastering Algebra - Advanced Level 1.4d Solving Quadratic Equations Using Completing-the-Square Method 
Hamilton Education Guides 86 
 
 Step 4 x x2 3
2
9
16
57
16
+ + = ; x +



=
3
4
57
16
2
 
 
 Step 5 x +



=
3
4
57
16
2
 ; x +


= ±
3
4
57
16
2
 ; x + = ±3
4
57
42
 ; x + = ±3
4
57
4
 therefore: 
 
 I. x + = +3
4
57
4
 ; x = −7 55
4
3
4
. ; x = −7 55 3
4
. ; x = 4 55
4
. ; x = 1138. 
 
 II. x + = −3
4
57
4
 ; x = − −7 55
4
3
4
. ; x = − −7 55 3
4
. ; x = −10 55
4
. ; x = −2 638. 
 and the solution set is { }−2 638 1138. , . . 
 Step 6 Check: Substitute x = 1138. and x = −2 638. in 2 3 6 02x x+ − = 
 I. Let x in x x= + − =1138 2 3 6 02. ; ( ) ( )2 1138 3 1138 6 02⋅ + × − =. .
?
 
 ; 2 59 341 6 0. .
?
+ − = ; 6 6 0− =
?
 ; 0 0= 
 II. Let x in x x= − + − =2 638 2 3 6 02. ; ( ) ( )2 2 638 3 2 638 6 02⋅ − + × − − =. .
?
 
 ; 1392 7 92 6 0. .
?
− − = ; 1392 1392 0. .
?
− = ; 0 0= 
 Step 7 Thus, the equation 2 3 6 02x x+ − = which is equal to x x2 15 3 0+ − =. can be 
 factored to ( )( )x x− + =1138 2 638 0. . 
Example 1.4-23 
 Solve the quadratic equation 2 6 7 02u u+ − = by completing the square. 
Solution: 
 Step 1 2 6 7 02u u+ − = ; 2 6 7 7 72u u+ − + = + ; 2 6 0 72u u+ + = ; 2 6 72u u+ = 
 
 Step 2 2 6 72u u+ = ; /
/
+
/
/
=
2
2
6
3
2
7
2
2u u ; u u2 3 7
2
+ = 
 
 Step 3 u u2 3 7
2
+ = ; u u2
2 2
3 3
2
7
2
3
2
+ + 



= + 



 ; u u2 3 9
4
7
2
9
4
+ + = + 
 
 ; ( ) ( )u u2 3 9
4
7 4 2 9
2 4
+ + =
⋅ + ⋅
⋅
 ; u u2 3 9
4
28 18
8
+ + =
+ ; u u2 3 9
4
46
8
+ + = 
 
 Step 4 u u2 3 9
4
46
8
+ + = ; u +



=
/ /
/
3
2
46
23
8
4
2
 ; u +



=
3
2
23
4
2
 
 
Mastering Algebra - Advanced Level 1.4d Solving Quadratic Equations Using Completing-the-Square Method 
Hamilton Education Guides 87 
 Step 5 u +



=
3
2
23
4
2
 ; u +


= ±
3
2
23
4
2
 ; u + = ±3
2
23
22
 ; u + = ±3
2
23
2
 therefore: 
 
 I. u + = +3
2
23
2
 ; u = − +
3
2
23
2
 ; u =
− +3 23
2
 ; u =
− +3 4 8
2
.
 ; u =
18
2
.
 ; u = 0 9. 
 
 II. u + = −3
2
23
2
 ; u = − −
3
2
23
2
 ; u =
− −3 23
2
 ; u =
− −3 4 8
2
.
 ; u =
−7 8
2
.
 ; u = −3 9. 
 and the solution set is { }−3 9 0 9. , . . 
 Step 6 Check: Substitute u = 0 9. and u = −39. in 2 6 7 02u u+ − = 
 I. Let u in u u= + − =0 9 2 6 7 02. ; ( )2 0 9 6 0 9 7 02⋅ + × − =. .
?
 ; 16 5 4 7 0. .
?
+ − = 
 ; 7 7 0− =
?
 ; 0 0= 
 II. Let u in u u= − + − =39 2 6 7 02. ; ( ) ( )2 39 6 39 7 02⋅ − + × − − =. .
?
 
 ; 2 15 2 234 7 0× − − =. .
?
 ; 30 4 234 7 0. .
?
− − = ; 7 7 0− =
?
 ; 0 0= 
 Step 7 Thus, the equation 2 6 7 02u u+ − = , which is the same as u u2 3 35 0+ − =. ,can 
 be factored to ( )( )u u− + =0 9 3 9 0. . . 
Additional Examples - Solving Quadratic Equations of the Form ax bx c2 0+ + = , where a 1 , by Completing the Square 
The following examples further illustrate how to solve quadratic equations, where the coefficient 
of the squared term is not equal to one, using Completing-the-Square method: 
Example 1.4-24 
Solve the quadratic equation 3 2 1 02x x+ − = using Completing-the-Square method. 
Solution: 
3 2 1 02x x+ − = ; 3 2 12x x+ = ; /
/
+ =
3
3
2
3
1
3
2x x ; x x2 2
3
1
3
+ = ; x x2
2 22
3
2
6
1
3
2
6
+ +
/
/




= +
/
/




 
 
; x x2
2 22
3
1
3
1
3
1
3
+ + 



= + 



 ; x x2 2
3
1
9
1
3
1
9
+ + = + ; x +



= +
1
3
1
3
1
9
2
 ; ( ) ( )x +



=
⋅ + ⋅
⋅
1
3
1 9 1 3
3 9
2
 
 
; x +



=
+1
3
9 3
27
2
 ; x +



=
1
3
12
27
2
 ; x +


= ±
1
3
12
27
2
 ; x + = ±1
3
12
27
 ; x + = ±1
3
0 44. 
 
; x + = ±0 33 0 67. . therefore: 
 
I. x + = +0 33 0 67. . ; x = −0 67 0 33. . ; x = 0 34. II. x + = −0 33 0 67. . ; x = − −0 67 0 33. . ; x = −1 
and the solution set is { }−1 0 34, . . 
Check: I. Let x in x x= − + − =1 3 2 1 02 ; ( ) ( )3 1 2 1 1 02⋅ − + ⋅ − − =
?
 ; 3 1 2 1 0⋅ − − =
?
 ; 3 3 0− =
?
 ; 0 0= 
Mastering Algebra - Advanced Level 1.4d Solving Quadratic Equations Using Completing-the-Square Method 
Hamilton Education Guides 88 
 II. Let x in x x= + − =0 34 3 2 1 02. ; ( )3 0 34 2 0 34 1 02⋅ + ⋅ − =. .
?
 ; 3 011 0 68 1 0⋅ + − =. .
?
 
 ; 0 33 0 68 1 0. .
?
+ − = ; 1 1 0− =
?
 ; 0 0= 
Therefore, the equation 3 2 1 02x x+ − = can be factored to ( )( )x x− + =0 34 1 0. . 
Example 1.4-25 
Solve the quadratic equation 3 12 4 02t t+ − = using Completing-the-Square method. 
Solution: 
3 12 4 02t t+ − = ; 3 12 42t t+ = ; /
/
+
/ /
/
=
3
3
12
4
3
4
3
2t x ; t t2 4 4
3
+ = ; t t2
2 2
4 4
2
2
4
3
4
2
2
+ +
/
/










= +
/
/










 
 
; t t2
2 2
4 2
1
4
3
2
1
+ + 



= + 



 ; t t2 4 4 4
3
4+ + = + ; ( )t + = +2 4
3
4
1
2 ; ( ) ( ) ( )t + = ⋅ + ⋅
⋅
2
4 1 4 3
3 1
2 
 
;( )t + = +2 4 12
3
2 ; ( )t + =2 16
3
2 ; ( )t + = ±2 16
3
2 ; t + = ±2 533. ; t + = ±2 2 31. therefore: 
 
I. t + = +2 2 31. ; t = −2 31 2. ; t = 0 31. II. t + = −2 2 31. ; t = − −2 31 2. ; t = −4 31. 
and the solution set is { }0 31 4 31. , .− . 
Check: I. Let t in t t= + − =0 31 3 12 4 02. ; ( ) ( )3 0 31 12 0 31 4 02⋅ + ⋅ − =. .
?
 ; 3 0 096 372 4 0⋅ − − =. .
?
 
 ; 0 288 372 4 0. .
?
+ − = ; 4 4 0− =
?
 ; 0 0= 
 II. Let t in t t= − + − =4 31 3 12 4 02. ; ( ) ( )3 4 31 12 4 31 4 02⋅ − + ⋅ − − =. .
?
 ; 3 18 57 5172 4 0⋅ − − =. .
?
 
 ; 5572 5172 4 0. .
?
− − = ; 5572 5572 0. .
?
− = ; 0 0= 
Therefore, the equation 3 12 4 02t t+ − = can be factored to ( )( )t t− + =0 31 4 31 0. . . 
Practice Problems - Solving Quadratic Equations of the Form ax bx c2 0+ + = , where a 1 , by Completing the Square 
Section 1.4d Case II Practice Problems - Solve the following quadratic equations using 
Completing-the-Square method. (Note that these problems are identical to the exercises given in 
Section 1.4b Case II.) 
1. 4 6 1 02u u+ + = 
 
2. 4 10 32w w+ = − 3. 6 4 2 02x x+ − = 
4. 15 3 142y y+ = − 
 
5. 2 5 3 02x x− + = 6. 2 02 2x xy y+ − = x is variable 
 
Mastering Algebra - Advanced Level 1.4e How to Choose the Best Factoring or Solution Method 
Hamilton Education Guides 89 
1.4e How to Choose the Best Factoring or Solution Method 
To factor polynomials and to solve quadratic equations a total of seven basic methods have been 
reviewed in Sections 1.3 and 1.4. Those methods are: 
1. The Greatest Common Factoring method 
2. The Grouping method 
3. The Trial and Error method 
4. Factoring methods for polynomials with square and cubed terms 
5. The Quadratic Formula method 
6. The Square Root Property method, and 
7. Completing-the-Square method 
The decision as to which one of the above methods is most suitable in factoring a polynomial or 
solving an equation is left to the student. For example, in some cases, using the Trial and Error 
method in solving a quadratic equation may be easier than using the Quadratic Formula or 
Completing-the-Square method. In certain cases, using the quadratic formula in solving a 
polynomial may be faster than the Grouping or the Trial and Error method. Note that the key in 
choosing the best and/or the easiest method is through solving many problems. After sufficient 
practice, students start to gain confidence on selection of one method over the other. 
Assumption - In many instances, the methods used in factoring polynomials (shown in Section 
1.3) can also be used in solving quadratic equations (shown in Section 1.4) by recognizing that 
the left hand side of the equation ax bx c2 0+ + = , namely ax bx c2 + + is a polynomial and can be 
factored as such, using polynomial factoring methods covered in Section 1.3. 
Note 1 - Any quadratic equation can be solved using the quadratic formula. Once the student has 
memorized the quadratic formula and has learned how to substitute the equivalent values of a , 
b , and c into the quadratic formula, then the next steps are merely the process of solving the 
quadratic equation using mathematical operations. 
Note 2 - The quadratic formula can be used as an alternative method in factoring polynomials of 
the form ax bx c2 + + as is stated in the above assumption. 
The following examples are solved using the seven factoring and solution methods shown above: 
Example 1.4-26 
Use different methods to solve the equation x2 25= . 
Solution: 
First Method: (The Trial and Error Method) 
Write the equation in the standard quadratic equation form ax bx c2 0+ + = , i.e., write x2 25= 
as x x2 0 25 0+ − = . To solve the given equation using the Trial and Error method we only 
consider the left hand side of the equation which is a second degree polynomial. Next, we 
need to obtain two numbers whose sum is 0 and whose product is −25 by constructing a 
table as shown below: 
 
Mastering Algebra - Advanced Level 1.4e How to Choose the Best Factoring or Solution Method 
Hamilton Education Guides 90 
Sum Product 
1 1 0− = ( )1 1 1⋅ − = − 
2 2 0− = ( )2 2 4⋅ − = − 
3 3 0− = ( )3 3 9⋅ − = − 
4 4 0− = ( )4 4 16⋅ − = − 
5 5 0− = ( )5 5 25⋅ − = − 
 
The last line contains the sum and the product of the two numbers that we need. Thus, 
x2 25= or x x2 0 25 0+ − = can be factored to ( )( )x x− + =5 5 0 . 
Check: ( )( )x x− + =5 5 0 ; ( )x x x x⋅ + ⋅ − ⋅ + ⋅ − =5 5 5 5 0 ; x x x2 5 5 25 0+ − − = ; ( )x x2 5 5 25 0+ − − = 
 ; x x2 0 25 0+ − = 
Second Method: (The Quadratic Formula Method) 
First, write the equation in the standard quadratic equation form ax bx c2 0+ + = , i.e., write 
x2 25= as x x2 0 25 0+ − = . Second, equate the coefficients of x x2 0 25 0+ − = with the 
standard quadratic equation by letting a = 1 , b = 0 , and c = −25 . Then, 
Given: x b b ac
a
=
− ± −2 4
2
 ; ( )x =
− ± − × × −
×
0 0 4 1 25
2 1
2
 ; x = ± +0 100
2
 ; x = ± 100
2
 ; x = ± 10
2
2
 
; x = ± 10
2
. Therefore: 
I. x = + / /
/
10
5
2
 ; x = 5
1
 ; x = 5 II. x = − / /
/
10
5
2
 ; x = − 5
1
 ; x = −5 
Check: I. Let x in x= =5 252 ; 5 252 =
?
 ; 25 25= 
 II. Let x in x= − =5 252 ; ( )− =5 252
?
 ; 25 25= 
Therefore, the equation x x2 0 25 0+ − = can be factored to ( )( )x x+ − =5 5 0 . 
Third Method: (The Square Root Property Method) 
Take the square root of both sides of the equation, i.e., write x2 25= as x2 25= ± ; 
x = ± 52 ; x = ±5 . Thus, x = +5 or x = −5 are the solution sets to the equation x2 25= which 
can be represented in its factorable form as ( )( )x x+ − =5 5 0 . 
Fourth Method: (Completing-the-Square Method) - Is not applicable. 
Note that from the above three methods using the Square Root Property method is the fastest and 
the easiest method to obtain the factored terms. The Trial and Error method is the second easiest 
method to use, followed by the Quadratic Formula method which is the most difficult way of 
obtaining the factored terms. 
Example 1.4-27 
Use different methods to solve the equation x x2 11 24 0+ + = . 
Solution: 
First Method: (The Trial and Error Method) 
To solve the given equation using the Trial and Error method we only consider the left hand 
Mastering Algebra - Advanced Level 1.4e How to Choose the Best Factoring or Solution Method 
Hamilton Education Guides 91 
side of the equation which is a second degree polynomial. Next, we need to obtain two 
numbers whose sum is 11 and whose product is 24 by constructing a table as shown below: 
Sum Product 
6 5 11+ = 6 5 30⋅ = 
7 4 11+ = 7 4 28⋅ = 
8 3 11+ = 8 3 24⋅ = 
9 2 11+ = 9 2 18⋅ = 
 
The third line contains the sum and the product of the two numbers that we need. Thus, 
x x2 11 24 0+ + = can be factored to ( )( )x x+ + =8 3 0 . 
Check: ( )( )x x+ + =8 3 0 ; x x x x⋅ + ⋅ + ⋅ + ⋅ =3 8 8 3 0 ; x x x2 3 8 24 0+ + + = ; ( )x x2 3 8 24 0+ + + = 
 ; x x2 11 24 0+ + = 
Second Method: (The Quadratic Formula Method) 
Given the standard quadratic equation ax bx c2 0+ + = , equate the coefficients of x x2 11 24 0+ + = 
with the standard quadratic equation by letting a = 1 , b = 11 , and c = 24 . Then, 
Given: x b b ac
a
=
− ± −2 4
2
 ; ( )x =
− ± − × ×
×
11 11 4 1 24
2 1
2
 ; x = − ± −11 121 96
2
 ; x = − ±11 25
2
 
; x = − ±11 5
2
2
 ; x = − ±11 5
2
. Therefore: 
I. x = − +11 5
2
 ; x = − /
/
6
3
2
 ; x = − 3
1
 ; x = −3 II. x = − −11 5
2
 ; x = − / /
/
16
8
2
 ; x = − 8
1
 ; x = −8 
Check: I. Let x in x x= − + + =3 11 24 02 ; ( ) ( )− + ⋅ − + =3 11 3 24 02
?
 ; 9 33 24 0− + =
?
 ; 33 33 0− =
?
 
 ; 0 0= 
 II. Let x in x x= − + + =8 11 24 02 ; ( ) ( )− + ⋅ − + =8 11 8 24 02
?
 ; 64 88 24 0− + =
?
 ; 88 88 0− =
?
 
 ; 0 0= 
Therefore, the equation x x2 11 24 0+ + = can be factored to ( )( )x x+ + =8 3 0 . 
Third Method: (Completing-the-SquareMethod) 
x x2 11 24 0+ + = ; x x2 11 24+ = − ; x x2
2 2
11 11
2
24 11
2
+ + 



= − + 



 ; x x2 11 121
4
24 121
4
+ + = − + 
; x +



= − +
11
2
24
1
121
4
2
 ; ( ) ( )x +



=
− ⋅ + ⋅
⋅
11
2
24 4 1 121
1 4
2
 ; x +



=
− +11
2
96 121
4
2
 ; x +



=
11
2
25
4
2
 
; x + = ±11
2
25
4
 ; x + = ±11
2
5
2
 
Therefore: I. x + = +11
2
5
2
 ; x = −5
2
11
2
 ; x = −5 11
2
 ; x = − /
/
6
3
2
 ; x = − 3
1
 ; x = −3 
 II. x + = −11
2
5
2
 ; x = − −5
2
11
2
 ; x = − −5 11
2
 ; x = − / /
/
16
8
2
 ; x = − 8
1
 ; x = −8 
Check: I. Let x in x x= − + + =3 11 24 02 ; ( ) ( )− + × − + =3 11 3 24 02
?
 ; 9 33 24 0− + =
?
 ; 0 0= 
Mastering Algebra - Advanced Level 1.4e How to Choose the Best Factoring or Solution Method 
Hamilton Education Guides 92 
 II. Let x in x x= − + + =8 11 24 02 ; ( ) ( )− + × − + =8 11 8 24 02
?
 ; 64 88 24 0− + =
?
 ; 0 0= 
Therefore, the equation x x2 11 24 0+ + = can be factored to ( )( )x x+ + =8 3 0 . 
Fourth Method: (The Square Root Property Method) - Is not applicable 
Note that from the above three methods using the Trial and Error method is the fastest and the 
easiest method to obtain the factored terms. Completing-the-Square method is the second easiest 
method to use, followed by the Quadratic Formula method which is the longest and most difficult 
way of obtaining the factored terms. 
Example 1.4-28 
Use different methods to solve the equation x x2 5 2 0+ + = . 
Solution: 
First Method: (The Trial and Error Method) 
To solve the given equation using the Trial and Error method we only consider the left hand 
side of the equation which is a second degree polynomial. Next, we need to obtain two 
numbers whose sum is 5 and whose product is 2 . However, after few trials, it becomes clear 
that such a combination of integer numbers is not possible to obtain. Hence, the Trial and 
Error method is not applicable to this particular example. 
Second Method: (The Quadratic Formula Method) 
Given the standard quadratic equation ax bx c2 0+ + = , equate the coefficients of 
x x2 5 2 0+ + = with the standard quadratic equation by letting a = 1 , b = 5 , and c = 2 . Then, 
Given: x b b ac
a
=
− ± −2 4
2
 ; ( )x =
− ± − × ×
×
5 5 4 1 2
2 1
2
 ; x = − ± −5 25 8
2
 ; x = − ±5 17
2
. Therefore: 
I. x = − +5 17
2
 ; x = − +5 412
2
. ; x = − 0 88
2
. ; x = −0 44. 
II. x = − −5 17
2
 ; x = − −5 412
2
. ; x = − 912
2
. ; x = −4 56. 
Check: I. Let x in x x= − + + =0 44 5 2 02. ; ( ) ( )− + ⋅ − + =0 44 5 0 44 2 02. .
?
 ; 0 2 2 2 2 0. .
?
− + = ; 2 2 2 2 0. .
?
− = 
 ; 0 0= 
 II. Let x in x x= − + + =4 56 5 2 02. ; ( ) ( )− + ⋅ − + =4 56 5 4 56 2 02. .
?
 ; 20 8 22 8 2 0. .
?
− + = 
 ; 22 8 22 8 0. .
?
− = ; 0 0= 
Therefore, the equation x x2 5 2 0+ + = can be factored to ( )( )x x+ + =0 44 4 56 0. . . 
Third Method: (Completing-the-Square Method) 
x x2 5 2 0+ + = ; x x2 5 2+ = − ; x x2
2 2
5 5
2
2 5
2
+ + 



= − + 



 ; x x2 5 25
4
2 25
4
+ + = − + ; x +



= − +
5
2
2
1
25
4
2
 
; ( ) ( )x +



=
− ⋅ + ⋅
⋅
5
2
2 4 1 25
1 4
2
 ; x +



=
− +5
2
8 25
4
2
 ; x +



=
5
2
17
4
2
 ; x + = ±5
2
17
4
 ; x + = ±5
2
17
2
 
Therefore: I. x + = +5
2
17
2
 ; x = −17
2
5
2
 ; x = −17 5
2
 ; x = −412 5
2
. ; x = − 0 88
2
. ; x = −0 44. 
 
Mastering Algebra - Advanced Level 1.4e How to Choose the Best Factoring or Solution Method 
Hamilton Education Guides 93 
 II. x + = −5
2
17
2
 ; x = − −17
2
5
2
 ; x = − −17 5
2
 ; x = − −412 5
2
. ; x = − 912
2
. ; x = −4 56. 
Check: I. Let x in x x= − + + =0 44 5 2 02. ; ( ) ( )− + ⋅ − + =0 44 5 0 44 2 02. .
?
 ; 0 2 2 2 2 0. .
?
− + = ; 2 2 2 2 0. .
?
− = 
 ; 0 0= 
 II. Let x in x x= − + + =4 56 5 2 02. ; ( ) ( )− + ⋅ − + =4 56 5 4 56 2 02. .
?
 ; 20 8 22 8 2 0. .
?
− + = 
 ; 22 8 22 8 0. .
?
− = ; 0 0= 
Therefore, the equation x x2 5 2 0+ + = can be factored to ( )( )x x+ + =0 44 4 56 0. . . 
Fourth Method: (The Square Root Property Method) - Is not applicable. 
Note that from the above two methods using the Quadratic Formula method may be the faster 
method, for some, than Completing-the-Square method. 
Practice Problems - How to Choose the Best Factoring or Solution Method 
Section 1.4e Practice Problems - Choose three methods to solve the following quadratic 
equations. State the degree of difficulty associated with each method you selected. 
1. x2 16= 2. x x2 7 3 0+ + = 3. ( )3 4 362x + = 4. x x2 11 30 0+ + = 
 
Mastering Algebra - Advanced Level 1.5a Introduction to Algebraic Fractions 
Hamilton Education Guides 94 
1.5 Algebraic Fractions 
 
In this section math operations involving algebraic fractions are reviewed. Algebraic fractions 
are introduced in Section 1.5a. The steps as to how algebraic fractions are simplified are 
addressed in Section 1.5b. Addition, subtraction, multiplication, and division of algebraic 
fractions are addressed in Section 1.5c. Math operations involving complex algebraic fractions 
are reviewed in Section 1.5d. 
1.5a Introduction to Algebraic Fractions 
Arithmetic/integer fractions are fractions where the numerator and the denominator are integer 
numbers. For example, 2
3
, 1
5
, − 3
8
, and 2
1
 are examples of arithmetic fractions. Algebraic 
fractions are fractions where the numerator or the denominator (or both) are variables. For 
example, a
3
, x
y
, 3
1x +
, and 1
x
 are examples of algebraic fractions. The concepts and procedures 
learned in simplifying, adding, subtracting, multiplying, and dividing arithmetic fractions can 
directly be applied to algebraic fractions. (The subject of arithmetic fractions has been addressed 
in detail in the “Mastering Fractions” book. Students are encouraged to review chapters 3 and 9 
for an overall review of the fractional operations.) In this section we will review the sign rules 
for fractions; division of algebraic fractions by zero; and equivalent algebraic fractions. 
A. Sign Rules For Fractions 
In division, we need to consider two signs. The sign of the numerator and the sign of the 
denominator. Thus, the sign rules for division are: 
 1. −
+
a
b
 = − a
b
 For example: −
+
8
2
 = − 8
2
 = − 4
1
 = −4 
 2. +
−
a
b
 = − a
b
 For example: +
−
8
2
 = − 8
2
 = − 4
1
 = −4 
 3. −
−
a
b
 = + a
b
 For example: −
−
8
2
 = + 8
2
 = + 4
1
 = +4 
 4. +
+
a
b
 = + a
b
 For example: +
+
8
2
 = + 8
2
 = + 4
1
 = +4 
In fractions, we need to consider three signs. The fractions sign itself, the numerator sign, and 
the denominator sign. Therefore, the sign rules for fractions are: 
1. + −
+
a
b
 = + −



a
b
 = − a
b
 2. + +
−
a
b
 = + −



a
b
 = − a
b
 
3. + −
−
a
b
 = + +



a
b
 = + a
b
 4. + +
+
a
b
 = + +



a
b
 = + a
b
 
5. − −
+
a
b
 = − −



a
b
 = + a
b
 6. − +
−
a
b
 = − −



a
b
 = + a
b
 
7. − −
−
a
b
 = − +



a
b
 = − a
b
 8. − +
+
a
b
 = − +



a
b
 = − a
b
 
For example: 
1. + −
+
8
2
 = + −



8
2
 = − 8
2
 = − 4
1
 = −4 2. + +
−
8
2
 = + −



8
2
 = − 8
2
 = − 4
1
 = −4 
Mastering Algebra - Advanced Level 1.5a Introduction to Algebraic Fractions 
Hamilton Education Guides 95 
3. + −
−
8
2
 = + +



8
2
 = + 8
2
 = + 4
1
 = +4 4. + +
+
8
2
 = + +



8
2
 = + 8
2
 = + 4
1
 = +4 
5. − −
+
8
2
 = − −



8
2
 = + 8
2
 = + 4
1
 = +4 6. − +
−
8
2
 = − −



8
2
 = + 8
2
 = + 4
1
 = +4 
7. − −
−
8
2
 = − +



8
2
 = − 8
2
 = − 4
1
 = −4 8. − +
+
8
2
 = − +



8
2
 = − 8
2
 = − 4
1
 = −4 
B. Divisionof Algebraic Fractions by Zero 
An algebraic fraction is an expression of the form 
A
B
 where A and B are polynomials 
Note that the denominator B in an algebraic fraction can not be equal to zero, since division by 
zero is not defined. For example, 
a. 12x
 is not defined when x = 0 because 1
02
 = 1
0
 is not defined. 
b. 2
1x +
 is not defined when x = −1 because 2
1 1− +
 = 2
0
 is not defined. 
c. a
a
−
−
2
2
 is not defined when a = 2 because 2 2
2 2
−
−
 = 0
0
 is not defined. 
d. 5
9
x
x −
 is not defined when x = 9 because 5 9
9 9
×
−
 = 45
0
 is not defined. 
e. x
x x
−
+ +
1
6 52
 = 
( ) ( )
x
x x
−
+ +
1
1 5
is not defined when x = −1 and x = −5 because 
when x = −1 
( ) ( )
− −
− + − +
1 1
1 1 1 5
 = −
×
2
0 4
 = − 2
0
 is not defined, and 
when x = −5 
( )( )
− −
− + − +
5 1
5 1 5 5
 = −
− ×
6
4 0
 = − 6
0
 is not defined. 
f. x
x x
+
+ −
5
2 152
 = 
( )( )
x
x x
+
− +
5
3 5
 = ( )
( )( )
/ + /
− / + /
x
x x
5
3 5
 = 1
3x −
 is not defined when x = 3 because 1
3 3−
 
 = 1
0
 is not defined. 
g. 3
3− −x
 is not defined when x = −3 because 
( )
3
3 3− − −
 = 3
3 3−
 = 3
0
 is not defined. 
h. 2 3
2 3
x
x
+
−
 is not defined when x = 3
2
 because 
/ ⋅
/
+
/ ⋅
/
−
2 3
2
3
2 3
2
3
 = 
3
1
3
3
1
3
+
−
 = 3 3
3 3
+
−
 = 6
0
 is not defined. 
i. x
x4 3−
 is not defined when x = 3
4
 because 
3
4
4 3
4
3⋅ −
 = 
3
4
4 3
4
3/ ⋅
/
−
 = 
3
4
3
1
3−
 = 
3
4
3 3−
 = 
3
4
0
 = 
3
4
0
1
 = 3 1
4 0
×
×
 
 = 3
0
 is not defined. 
Mastering Algebra - Advanced Level 1.5a Introduction to Algebraic Fractions 
Hamilton Education Guides 96 
C. Equivalent Algebraic Fractions 
When the numerator and the denominator of an algebraic fraction is multiplied by the same 
number, sign, or a variable the new algebraic fraction is said to be equivalent to the original 
algebraic fraction. For example, the following algebraic fractions are equivalent to one another: 
a. a
b
 = 3
3
a
b
 = 100
100
a
b
 = 5
5
a
b
 = −
−
a
b
 
b. 1+ x
x
 = 2 2
2
+ x
x
 = a ax
ax
+ 
c. x − 3
4
 = 5 15
20
x − = 3 9
12
x − = ax a
a
− 3
4
 = ( )− −
−
x 3
4
 = 3
4
−
−
x 
d. x
x
2
21+
 = 5
5 5
2
2
x
x+
 = 
2
2
1
2
1
2
2
1
x
x
+
 = x y
y x y
2
2+
 
e. a b
a
−
3
 = 2 2
6
a b
a
− = 10 10
30
a b
a
− = ax bx
ax
−
3
 
f. 5
x y−
 = 
( )
−
− −
5
x y
 = −
−
5
y x
 = 10
2 2x y−
 = 52 2
xy
x y xy−
 
g. x
x
−
−
3
2 7
 = ( )
( )
− −
− −
x
x
3
2 7
 = 3
7 2
−
−
x
x
 = ( )
( )
2 3
2 2 7
⋅ −
⋅ −
x
x
 = 2 6
4 14
x
x
−
−
 = ( )
( )
− ⋅ −
− ⋅ −
3 3
3 2 7
x
x
 = 9 3
21 6
−
−
x
x
 
h. −
−
2
1x
 = 
( )
−
−
− −
2
1x
 = 2
1− x
 = −
−
6
3 3x
 = 6
3 3− x
 
i. 1
1
−
−
x
y
 = ( )
( )
− −
− −
1
1
x
y
 = x
y
−
−
1
1
 = ( )
( )
2 1
2 1
⋅ −
⋅ −
x
y
 = 2 2
2 2
−
−
x
y
 
Practice Problems - Introduction to Algebraic Fractions 
A. Section 1.5a Practice Problems - Write the correct sign for the following fractions. 
1. −
−
2
5
 = 2. + −
−
3
6
 = 3. − −
−
8
4
 = 
4. 10
2−
 = 5. −
−
5
15
 = 6. + −8
6
 = 
B. Section 1.5a Practice Problems - State the value(s) of the variable for which the following 
fractions are not defined. 
1. 3
1x −
 2. x
x
−
−
5
5
 3. − 1
x
 
4. x
x +10
 5. 2
3 5
x
x −
 6. 5 2
7
x
x
−
−
 
C. Section 1.5a Practice Problems - State which of the following algebraic fractions are equivalent. 
1. 2
3
4
9
x
y
x
y
=
?
 2. 3 1
2
9 3
6
x
x
x
x
+
=
+? 3. 2 2
a b b a−
=−
−
?
 
4. x
x
x
x
−
+
=
−
−
5
1
5
1
?
 5. −
−
=
−
a
a
a
a1 1
?
 6. 3 3−
−
=
+x
x
x
x
?
 
 
Mastering Algebra - Advanced Level 1.5b Simplifying Algebraic Fractions to Lower Terms 
Hamilton Education Guides 97 
1.5b Simplifying Algebraic Fractions to Lower Terms 
In dealing with integer fractions we learned that integer (arithmetic) fractions are reduced to their 
lowest terms by dividing both the numerator and the denominator by their common terms. For 
example, the integer fraction 14
7
 is simplified to its lowest term by dividing both the numerator 
and the denominator by 7 , which is common to both, i.e., 14
7
 = / ⋅
/
7 2
7
 = 2
1
 = 2 . The same 
principle holds true when simplifying algebraic fractions. Algebraic fractions are simplified 
using the following steps: 
Step 1 Factor both the numerator and the denominator completely (see Sections 1.3 and 1.4). 
Step 2 Simplify the algebraic fraction by eliminating the common terms in both the 
numerator and the denominator. 
Examples with Steps 
The following examples show the steps as to how algebraic fractions are simplified to their 
lowest terms: 
Example 1.5-1 
 2 7 15
25
2
2
y y
y
− −
−
 = 
Solution: 
 Step 1 2 7 15
25
2
2
y y
y
− −
−
 = ( )( )2 3 5
52 2
y y
y
+ −
−
 = ( )( )( )( )
2 3 5
5 5
y y
y y
+ −
− +
 
 
 Step 2 ( )( )( )( )
2 3 5
5 5
y y
y y
+ −
− +
 = ( )( )( )( )
2 3 5
5 5
y y
y y
+ / − /
/ − / +
 = 2 35
y
y
+
+
 
Example 1.5-2 
 x x
x x x
3
3 22 3
−
− −
 = 
Solution: 
 Step 1 x x
x x x
3
3 22 3
−
− −
 = 
( )
( )
x x
x x x
2
2
1
2 3
−
− −
 = ( )( )( )( )
x x x
x x x
− +
+ −
1 1
1 3
 
 
 Step 2 ( )( )( )( )
x x x
x x x
− +
+ −
1 1
1 3
 = ( )( )( )( )
/ − / + /
/ / + / −
x x x
x x x
1 1
1 3
 = xx
−
−
1
3 
Additional Examples - Simplifying Algebraic Fractions to Lower Terms 
The following examples further illustrate how to simplify algebraic fractions to their lowest terms: 
Example 1.5-3 
x x
x x
2
2
5
2 15
+
+ −
 = ( )( )( )
x x
x x
+
− +
5
3 5
 = ( )( )( )
x x
x x
/ + /
− / + /
5
3 5
 = xx − 3 
Mastering Algebra - Advanced Level 1.5b Simplifying Algebraic Fractions to Lower Terms 
Hamilton Education Guides 98 
Example 1.5-4 
x
x
2 1
4 4
−
+
 = ( )( )( )
x x
x
− +
+
1 1
4 1
 = ( )( )( )
x x
x
− / + /
/ + /
1 1
4 1
 = x −14 
Example 1.5-5 
7
14 72
x
x x− +
 = ( )
7
7 2 1
x
x x− +
 = ( )
/ /
/ / − +
7
7 2 1
x
x x
 = 1
2 1− +x
 = −
−
1
2 1x 
Example 1.5-6 
−
−
−
5 3
3 5
x
x
 = ( )− −
−
5 3
3 5
x
x
 = 3 5
3 5
−
−
x
x
 = / − / /
/ − / /
3 5
3 5
x
x
 = 1
1
 = 1 
Example 1.5-7 
6 62
2 2
a ab
a b
−
−
 = ( )( )( )
6a a b
a b a b
−
− +
 = ( )( )( )
6a a b
a b a b
/ − /
/ − / +
 = 6aa b+ 
Example 1.5-8 
3 9
6
2
3 2
x x
x x x
+
+ −
 = ( )( )
3 3
62
x x
x x x
+
+ −
 = ( )( )( )
3 3
3 2
x x
x x x
+
+ −
 = ( )( )( )
3 3
3 2
/ / + /
/ / + / −
x x
x x x
 = 3 2x − 
Example 1.5-9 
6 1
3 2 1
2
2
x x
x x
+ −
+ −
 = ( )( )( )( )
2 1 3 1
3 1 1
x x
x x
+ −
− +
 = ( )( )( )( )
2 1 3 1
3 1 1
x x
x x
+ / / − /
/ / − / +
 = 2 11
x
x
+
+
 
Example 1.5-10 
( )u
u u
+
+ +
1
2 3 1
2
2 = 
( )( )
( )( )
u u
u u
+ +
+ +
1 1
1 2 1
 = ( )( )( )( )
/ + / +
/ + / +
u u
u u
1 1
1 2 1
 = uu
+
+
1
2 1 
Example 1.5-11 
t
t t
2
2
9
2 15
−
− −
 = t
t t
2 2
2
3
2 15
−
− −
 = ( )( )( )( )
t t
t t
− +
+ −
3 3
3 5
 = ( )( )( )( )
t t
t t
− / + /
/ + / −
3 3
3 5
 = tt
−
−
3
5 
Example 1.5-12 
6 7 3
3
2
3 2
y y
y y
+ −
− +
 = ( )( )
( )
2 3 3 1
3 12
y y
y y
+ −
− −
 = ( )( )
( )
2 3 3 1
3 12
y y
y y
+ / / − /
− / / − /
 = 2 32
y
y
+
−
 = − +2 32
y
y
 
Practice Problems - Simplifying Algebraic Fractions to Lower Terms 
Section 1.5b Practice Problems - Simplify the following algebraic fractions to their lowest terms: 
1. x y z
xy z
2 2 5
3 2−
 = 
 
2. −
−
3
9
2 3
2
a bc
ab c
 = 3. 
1 2
1 2
+
−
m
m
 = 
4. 2
10
3
2
uvw
u v
 = 5. y
y y
2
2
4
6
−
− −
 = 6. x x
x
3 2
2
3
9
−
−
 = 
 
Mastering Algebra - Advanced Level 1.5c Math Operations Involving Algebraic Fractions 
Hamilton Education Guides 99 
1.5c Math Operations Involving Algebraic Fractions 
In this section addition, subtraction, multiplication, and division of algebraic fractions (Cases I 
through IV) are reviewed. 
Case I Addition and Subtraction of AlgebraicFractions with Common Denominators 
Algebraic fractions with common denominators are added and subtracted using the following 
steps: 
Step 1 Write the common denominator. Add or subtract the numerators. 
Step 2 Simplify the algebraic fraction to its lowest term. 
Examples with Steps 
The following examples show the steps as to how algebraic fractions with common denominators 
are added and subtracted: 
Example 1.5-13 
 ( )( ) ( )( )
3 5 5
3 1
3 4 2
3 1
2 2x x
x x
x x
x x
+ +
+ −
−
+ +
+ −
 = 
Solution: 
 Step 1 ( )( ) ( )( )
3 5 5
3 1
3 4 2
3 1
2 2x x
x x
x x
x x
+ +
+ −
−
+ +
+ −
 = 
( )
( )( )
3 5 5 3 4 2
3 1
2 2x x x x
x x
+ + − + +
+ −
 
 
 Step 2 
( )
( )( )
3 5 5 3 4 2
3 1
2 2x x x x
x x
+ + − + +
+ −
 = ( )( )
3 5 5 3 4 2
3 1
2 2x x x x
x x
+ + − − −
+ −
 
 
 = 
( ) ( ) ( )
( )( )
3 3 5 4 5 2
3 1
2 2x x x x
x x
− + − + −
+ −
 = ( ) ( )( )( )
3 3 5 4 3
3 1
2− + − +
+ −
x x
x x
 = ( )( )
0 3
3 1
2x x
x x
+ +
+ −
 
 
 = ( )( )( )
/ + /
/ + / −
x
x x
3
3 1
 = 1 1x − 
Example 1.5-14 
 3
2
2
2
2
22 3 2 3 2 3
a b
a b
a b
a b
a b
a b
+
+
−
+
− = 
Solution: 
 Step 1 3
2
2
2
2
22 3 2 3 2 3
a b
a b
a b
a b
a b
a b
+
+
−
+
− = ( ) ( )3 2 2
2 2 3
a b a b a b
a b
+ + − + −
 
 
 Step 2 ( ) ( )3 2 2
2 2 3
a b a b a b
a b
+ + − + −
 = ( ) ( )3 2 2
2 2 3
a a a b b b
a b
+ + + − −
 = ( ) ( )3 2 1 1 1 2
2 2 3
+ + + − −a b
a b
 
 
Mastering Algebra - Advanced Level 1.5c Math Operations Involving Algebraic Fractions 
Hamilton Education Guides 100 
 = 6 2
2 2 3
a b
a b
− = ( )/ −
/
2 3
2 2 3
a b
a b
 = 3 2 3
a b
a b
− 
Additional Examples - Addition and Subtraction of Algebraic Fractions with Common Denominators 
The following examples further illustrate how to add or subtract algebraic fractions with common 
denominators: 
Example 1.5-15 
( )( ) ( )( )
x
x x
x
x x
−
+ −
−
−
+ −
3
4 4
5
4 4
 = ( )( )( )
x x
x x
− − −
+ −
3 5
4 4
 = ( )( )
x x
x x
− − +
+ −
3 5
4 4
 = ( ) ( )( )( )
x x
x x
− + − +
+ −
3 5
4 4
 = ( )( )( )
1 1 2
4 4
− +
+ −
x
x x
 
 
= ( )( )
0 2
4 4
x
x x
+
+ −
 = ( )( )
2
4 4x x+ − 
Example 1.5-16 
x
x
x
x
+
−
+
+
−
2
3
3
3
 = x x
x
+ + +
−
2 3
3
 = ( ) ( )x x
x
+ + +
−
2 3
3
 = ( )1 1 5
3
+ +
−
x
x
 = 2 53
x
x
+
−
 
Example 1.5-17 
( )( ) ( )( )
x x
x x
x x
x x
2 23 2
1 5
2 1
1 5
+ +
+ −
−
+ +
+ −
 = 
( )
( )( )
x x x x
x x
2 23 2 2 1
1 5
+ + − + +
+ −
 = ( )( )
x x x x
x x
2 23 2 2 1
1 5
+ + − − −
+ −
 
 
= 
( ) ( ) ( )
( )( )
x x x x
x x
2 2 3 2 2 1
1 5
− + − + −
+ −
 = ( ) ( )( )( )
1 1 3 2 1
1 5
2− + − +
+ −
x x
x x
 = ( )( )
0 1
1 5
2x x
x x
+ +
+ −
 = ( )( )( )
/ + /
/ + / −
x
x x
1
1 5
 = 1 5x − 
Example 1.5-18 
3
3
2 1
3
3 2
3
x
x
x
x
x
x+
+
−
+
+
+
+
 = 3 2 1 3 2
3
x x x
x
+ − + +
+
 = ( ) ( )3 2 3 1 2
3
x x x
x
+ + + − +
+
 = ( )3 2 3 1
3
+ + +
+
x
x
 = 8 13
x
x
+
+
 
Example 1.5-19 
x x
x
x
x
x
x
2 22 4
2
1
2
3
2
+ +
+
−
−
+
−
−
+
 = 
( ) ( )x x x x
x
2 22 4 1 3
2
+ + − − − −
+
 = x x x x
x
2 22 4 1 3
2
+ + − + − +
+
 
 
= 
( ) ( ) ( )x x x x
x
2 2 2 4 1 3
2
− + − + + +
+
 = ( ) ( )1 1 2 1 8
2
2− + − +
+
x x
x
 = 0 8
2
2x x
x
+ +
+
 = xx
+
+
8
2 
Practice Problems - Addition and Subtraction of Algebraic Fractions with Common Denominators 
Section 1.5c Case I Practice Problems - Add or subtract the following algebraic fractions. Reduce 
the answer to its lowest term. 
1. x
7
3
7
+ = 2. 8 7
a b a b+
−
+
 = 3. 3 1
2
4 1
2
3
2
x
y
x
y y
+
+
+
+ = 
4. 4
2
8
2
x
x x−
−
−
 = 5. 15
5
5
5
a
a b
b
a b+
−
−
+
 = 6. 6
3
5
32 2 2 2
x
x y
y
x y
+ = 
 
Mastering Algebra - Advanced Level 1.5c Math Operations Involving Algebraic Fractions 
Hamilton Education Guides 101 
Case II Addition and Subtraction of Algebraic Fractions without Common Denominators 
Algebraic fractions without common denominators are solved using the following steps: 
Step 1 Obtain a common denominator by multiplying the denominators of the first and 
second fractions by one another. Cross multiply the numerator of the first fraction 
with the denominator of the second fraction. Cross multiply the numerator of the 
second fraction with the denominator of the first fraction. Add or subtract the two 
products to each other. 
Step 2 Simplify the algebraic fraction to its lowest term. 
Examples with Steps 
The following examples show the steps as to how algebraic fractions without common 
denominators are added and subtracted: 
Example 1.5-20 
 2
2
4
3x x−
+
+
 = 
Solution: 
 Step 1 2
2
4
3x x−
+
+
 = 
( )[ ] ( )[ ]
( ) ( )
2 3 4 2
2 3
⋅ + + ⋅ −
− ⋅ +
x x
x x
 = ( )( )
2 6 4 8
2 3
x x
x x
+ + −
− +
 
 
 Step 2 ( )( )
2 6 4 8
2 3
x x
x x
+ + −
− +
 = ( ) ( )( )( )
2 4 8 6
2 3
x x
x x
+ + − +
− +
 = ( )( )
6 2
2 3
x
x x
−
− +
 = 
( )
( ) ( )
2 3 1
2 3
x
x x
−
− +
 
Example 1.5-21 
 m
m
m
m
+
−
−
2
1
 = 
Solution: 
 Step 1 m
m
m
m
+
−
−
2
1
 = 
( ) ( )[ ] ( )
( )
m m m m
m m
+ ⋅ − − ⋅
⋅ −
2 1
1
 = ( )
m m m m
m m
2 22 2
1
− + − −
−
 
 
 Step 2 ( )
m m m m
m m
2 22 2
1
− + − −
−
 = 
( ) ( )
( )
m m m m
m m
2 2 2 2
1
− + − −
−
 = ( )
m
m m
−
−
2
1 
Additional Examples - Addition and Subtraction of Algebraic Fractions without Common Denominators 
The following examples further illustrate how to add or subtract algebraic fractions without 
common denominators: 
Example 1.5-22 
7 3
32 2 2x y xy
− = 
( ) ( )7 3 3
3
2 2 2
2 2 2
⋅ − ⋅
⋅
xy x y
x y xy
 = 21 3
3
2 2 2
3 4
xy x y
x y
− = ( )3 7
3
2
3 4
xy x
x y
−
 = ( )/ / −
/
/
/ /
3 7
3
2
3
2
4
2
xy x
x
x
y
y
 = 72 2
− x
x y
 
 
Mastering Algebra - Advanced Level 1.5c Math Operations Involving Algebraic Fractions 
Hamilton Education Guides 102 
Example 1.5-23 
x y
x y
x
y
+
−
+ = 
( )[ ] ( )[ ]
( )
x y y x x y
y x y
+ ⋅ + ⋅ −
⋅ −
 = ( )
xy y x xy
y x y
+ + −
−
2 2
 = ( )( )
/ / − / / + +
−
xy xy x y
y x y
2 2
 = ( )
x y
y x y
2 2+
−
 
Example 1.5-24 
3 2
42a a
+ = 
( ) ( )3 4 2
4
2
2
⋅ + ⋅
⋅
a a
a a
 = 12 2
4
2
3
a a
a
+ = ( )2 6
4 3
a a
a
+
 = ( )/ / +
/ /
2 6
4
2
3
2
a a
a
a
 = 6
2 2
+ a
a
 
Example 1.5-25 
x
yz
y
xz
z
xy
+ + = x
yz
y
xz
z
xy
+





 + = ( ) ( )x xz y yz
yz xz
z
xy
⋅ + ⋅
⋅





 + = x z y z
xyz
z
xy
2 2
2
+




 + = 
( )/ +
+/
z x y
xy z
z
z
xy
2 2
2 
 
= x y
xyz
z
xy
2 2+
+ = 
( )[ ] ( )[ ]xy x y z xyz
xyz xy
⋅ + + ⋅
⋅
2 2
 = x y xy xyz
x y z
3 3 2
2 2
+ + = 
( )/ / + +
/ /
xy x y z
x
x
y
y
z
2 2 2
2 2 = 
x y z
xyz
2 2 2+ + 
Example 1.5-26 
3 1
2
2 3
3
a
a
a
a
−
−
− = 
( )[ ] ( )[ ]3 1 3 2 3 2
2 3
a a a a
a a
− ⋅ − − ⋅
⋅
 = 9 3 4 6
6
2 2
2
a a a a
a
− − + = 
( ) ( )9 4 6 3
6
2 2
2
a a a a
a
− + −
 
 
= 5 3
6
2
2
a a
a
+ = ( )/ +/
a a
a
a
5 3
6 2
 = 5 36
a
a
+ 
Example 1.5-27 
x
x
x
x
−
+
−
−
+
1
2
2
1
 = 
( ) ( )[ ] ( ) ( )[ ]
( ) ( )
x x x x
x x
− ⋅ + − − ⋅ +
+ ⋅ +
1 1 2 2
2 1
 = 
( ) ( )
( )( )
x x x x x x
x x
2 21 2 2 4
2 1
+ / − / − − − / / + / / −
+ +
 = ( )( )
x x
x x
2 21 4
2 1
− − +
+ +
 
 
= 
( ) ( )
( )( )
x x
x x
2 2 1 4
2 1
− + − +
+ +
 = 
( ) ( )
( )( )
x x
x x
/ /− + − +
+ +
2 2 1 4
2 1
 = ( )( )
3
2 1x x+ + 
Example 1.5-28 
m
m n
n
m n−
−
+
 = 
( )[ ] ( )[ ]
( ) ( )
m m n n m n
m n m n
⋅ + − ⋅ −
− ⋅ +
 = m mn mn n
m mn mn n
2 2
2 2
+ − +
+ − −
 = m mn mn n
m mn mn n
2 2
2 2
+ / / − / / +
+ / / − / / −
 = m n
m n
2 2
2 2
+
−
 
Practice Problems - Addition and Subtraction of Algebraic Fractions without Common Denominators 
Section 1.5c Case II Practice Problems - Add or subtract the following algebraic fractions. 
Reduce the answer to its lowest term. 
1. 3
4
5
22 3x x
+ = 2. x
x x+
−
−4
2
1
 = 3. a b
a b
a
b
−
+
− = 
4. x
x
x
x
2
3
5
5+
+
−
 = 5. 
1
4
2
2 2 2x y z xy z− = 6. 3
1
2
1
5
x x x+
+
−
− = 
Mastering Algebra - Advanced Level 1.5c Math Operations Involving Algebraic Fractions 
Hamilton Education Guides 103 
Case III Multiplication of Algebraic Fractions 
To multiply algebraic fractions by one another we only review simple cases of algebraic fractions 
where the numerator and the denominator are mostly monomials. The more difficult algebraic 
expressions where the terms in the numerator and/or the denominator are polynomials and need 
to be factored first before reducing the algebraic fraction to lower terms are addressed in Chapter 
5 of the “Mastering Algebra – Intermediate Level”. Algebraic fractions are multiplied by one 
another using the following steps: 
Step 1 Write the algebraic expression in fraction form, i.e., write x or u v w2 2 3 as x
1
 and 
u v w2 2 3
1
, respectively. 
Step 2 Multiply the numerator and the denominator of the algebraic fraction terms by one 
another. Simplify the product to its lowest term. 
Examples with Steps 
The following examples show the steps as to algebraic fractions are multiplied by one another: 
Example 1.5-29 
 u v w
uv
w
v w
2 2
2 2 3
1
⋅ ⋅ = 
Solution: 
 Step 1 u v w
uv
w
v w
2 2
2 2 3
1
⋅ ⋅ = u v w
uv
w
v w
2 2
2 2 31
1
⋅ ⋅ 
 
 Step 2 u v w
uv
w
v w
2 2
2 2 31
1
⋅ ⋅ = u v w w
uv v w
2 2
2 2 3
1
1
⋅ ⋅
⋅ ⋅
 = u v w
uv w
2 2 2
4 3 = 
u
u
v w
u v
v
w
w
/ / /
/ /
2 2 2
4
2
3 = 
u
v w2
 
Example 1.5-30 
 ( ) ( )x x
x
x
x
x− ⋅
−
⋅
+
⋅3
9
3
2
82
2 = 
Solution: 
 Step 1 ( ) ( )x x
x
x
x
x− ⋅
−
⋅
+
⋅3
9
3
2
82
2 = ( ) ( )x x
x
x
x
x−
⋅
−
⋅
+
⋅
3
1 9
3
2
8
12
2
 
 
 Step 2 ( ) ( )x x
x
x
x
x−
⋅
−
⋅
+
⋅
3
1 9
3
2
8
12
2
 = ( ) ( )( )
x x x x
x x
− ⋅ ⋅ + ⋅
⋅ − ⋅
3 3 8
1 9 2
2
2 = 
( )( )
( )( )
8 3 3
2 3 3
3x x x
x x x
− +
− +
 
 
 = ( )( )( )( )
8 3 3
2 3 3
3x x x
x x x
/ − / / + /
/ − / / + /
 = 8
2
3x
x
 = /
/ /
/8
4
2
3
2
x
x
x
 = 4
1
2x = 4 2x 
Mastering Algebra - Advanced Level 1.5c Math Operations Involving Algebraic Fractions 
Hamilton Education Guides 104 
Additional Examples - Multiplication of Algebraic Fractions 
The following examples further illustrate how to multiply algebraic fractions: 
Example 1.5-31 
ab c a
b c
2
2 2⋅ = 
ab c a
b c
2
2 21
⋅ = ab c a
b c
2
2 21
⋅
⋅
 = a b c
b c
2 2
2 2 = 
a b c
b c
c
2 2
2 2
/
/ /
/ = ac
2
 
Example 1.5-32 
( )( )
x x
x x
+
⋅
−
+ −
1
2
4
1 2
2
 = ( )( )( )( )
x x x
x x
+
⋅
+ −
+ −
1
2
2 2
1 2
 = 
( ) ( )( )[ ]
( )( )[ ]
x x x
x x
+ ⋅ − +
⋅ + −
1 2 2
2 1 2
 = 
( ) ( )( )[ ]
( )( )[ ]
/ + / ⋅ / − / +
⋅ / + / / − /
x x x
x x
1 2 2
2 1 2
 = x + 22 
Example 1.5-33 
2
42 2 2
2 3
x y z
x z
yz
⋅ = 2
4
2 3
2 2 2
⋅
⋅
x z
x y z yz
 = 2
4
2 3
2 3 3
x z
x y z
 = /
/
/ /
/ /
2
4
2
2 3
2 3 3
x z
x y z
 = 1
2 3y
 
Example 1.5-34 
1
2
3
52
2
a
a a
⋅ ⋅ = 1 3
2 5
2
2
⋅ ⋅
⋅ ⋅
a a
a
 = 3
10
3
2
a
a
 = 3
10
3
2
a
a
a
/
/ = 
3
10
a 
Example 1.5-35 
uv
w
w
u v u
2
3
2
2 2
1
⋅ ⋅ = uv w
w u v u
2 2
3 2 2
1⋅ ⋅
⋅ ⋅
 = uv w
u v w
2 2
3 2 3 = 
/ / /
/ / /
uv w
u
u
v w
w
2 2
3
2
2 3 = 
1
2u w
 
Example 1.5-36 
xyz
x y z
xy
z
2
2 2 3
1
⋅ ⋅





 = 
xyz
x y z
xy
z
2
2 2 31
1
⋅ ⋅





 = 
xyz xy
x y z z
2
2 2 31
1
⋅
⋅
⋅





 = 
xyz xy
x y z
2
2 2 41
⋅





 = 
xyz xy
x y z
2
2 2 41
⋅ 
 
= xyz xy
x y z
2
2 2 41
⋅
⋅
 = x y z
x y z
2 2 2
2 2 4 = 
x y z
x y z
z
/ / /
/ / /
2 2 2
2 2 4
2
 = 
1
2z
 
Practice Problems - Multiplication of Algebraic Fractions 
Section 1.5c Case III Practice Problems - Multiply the following algebraic fractions by one 
another. Simplify the answer to its lowest term. 
1. 1
2
2 2
xy
x y
⋅ = 2. 2 1
8
2
3
a
a a
⋅ = 3. xyz
x y z
x
y
⋅ ⋅
1
2 2 2
2
 = 
4. 5
15
12 2 3
2 4
u v
uv
uv
v u
⋅ ⋅ = 5. 8
2 1
43 2
x
x x
⋅ ⋅ = 6. x
x
x
x
−
+
⋅
−
⋅
−
4
2
4
2
1
4
2
 = 
 
Mastering Algebra - Advanced Level 1.5c Math Operations Involving Algebraic Fractions 
Hamilton Education Guides 105 
Case IV Division of Algebraic Fractions 
To divide algebraic fractions by one another we only review simple cases of algebraic factions 
where the numerator and the denominator are mostly monomials. The more difficult algebraic 
expressions where the terms in the numerator and/or the denominator are polynomials and need 
to be factored first before reducing the algebraic fraction to lower terms are addressed in Chapter 
5 of the “Mastering Algebra – Intermediate Level”. Algebraic fractions are divided by one 
another using the following steps: 
Step 1 Write the algebraic expression in fraction form, i.e., write x or u v w2 2 3 as x
1
 and 
u v w2 2 3
1
, respectively. 
Step 2 Invert the second fraction and change the division sign to a multiplication sign. 
Step 3 Multiply the numerator and the denominator of the algebraic fraction terms by one 
another. Simplify the product to its lowest term, if possible. 
Examples with Steps 
The following examples show the steps as to algebraic fractions are divided by one another: 
Example 1.5-37 
 1 33 3 2 2x y z
z
x y
÷ = 
Solution: 
 Step 1 Not Applicable 
 
 Step 2 1 33 3 2 2x y z
z
x y
÷ = 1
33 3
2 2
x y z
x y
z
⋅ 
 
 Step 3 1
33 3
2 2
x y z
x y
z
⋅ = 1
3
2 2
3 3
⋅
⋅
x y
x y z z
 = x y
x y z
2 2
3 3 23
 = x y
x
x
y
y
z
/ /
/ /
2 2
3 3 23
 = 1
3 2xyz
 
Example 1.5-38 
 a b ab
ab
2 2
33
÷ = 
Solution: 
 Step 1 a b ab
ab
2 2
33
÷ = a b ab
ab
2 2
31 3
÷ 
 
 Step 2 a b ab
ab
2 2
31 3
÷ = a b ab
ab
2 2 3
1
3
⋅ 
 
 Step 3 a b ab
ab
2 2 3
1
3
⋅ = a b ab
ab
2 2 33
1
⋅
⋅
 = 3
3 5a b
ab
 = 3
1
2 4a b = 3 2 4a b 
Mastering Algebra - Advanced Level 1.5c Math Operations Involving Algebraic Fractions 
Hamilton Education Guides 106 
Example 1.5-39 
 3 1 23 2
3x
x x
x÷ ÷



 = 
Solution: 
 Step 1 3 1 23 2
3x
x x
x÷ ÷



 = 3 1 2
13 2
3x
x x
x
÷ ÷





 
 
 Step 2 3 1 2
13 2
3x
x x
x
÷ ÷





 = 
3 1 1
23 2 3
x
x x x
÷ ⋅




 = 3 1 1
23 2 3
x
x x x
÷
⋅
⋅





 = 3 1
23 5
x
x x
÷ = 3 2
13
5x
x
x
⋅ 
 
 Step 3 3 2
13
5x
x
x
⋅ = 3 2
1
5
3
x x
x
⋅
⋅
 = 6
6
3
x
x
 = 6
1
3x = 6 3x 
Additional Examples - Division of Algebraic Fractions 
The following examples further illustrate how to divide algebraic fractions by one another: 
Example 1.5-40 
x y z
xy
x y
yz
2 2 3
2
2 2
÷ = x y z
xy
yz
x y
2 2 3
2 2 2⋅ = 
x y z yz
xy x y
2 2 3
2 2 2
⋅
⋅
 = x y z
x y
2 3 4
3 4 = 
z
xy
4
 
Example 1.5-41 
a b
bc
ab
2 2
2÷ = a b
bc
ab2 2 2
1
÷ = a b
bc ab
2 2
2
1
⋅ = a b
bc ab
2 2
2
1⋅
⋅
 = a b
ab c
2 2
3 = 
a
bc 
Example 1.5-42 
3 9
2 2 3u v
u
u v
÷ = 3
92 2
3
u v
u v
u
⋅ = 3
9
3
2 2
⋅
⋅
u v
u v u
 = 3
9
3
3 2
u v
u v
 = / /
/
/
/ /
3
9
3
3
3 2
u v
u v
v
 = 13v 
Example 1.5-43 
xyz x y z
xy
3
2 2
2
÷ = xyz x y z
xy
3 2 2
1 2
÷ = xyz xy
x y z
3
2 21
2
⋅ = xyz xy
x y z
3
2 2
2
1
⋅
⋅
 = 2
2 2 3
2 2
x y z
x y z
 = 2
1
2z = 2 2z 
Example 1.5-44 
a b c
ab
a b
c
3 2
2
2
3÷ = 
a b c
ab
c
a b
3 2
2
3
2⋅ = 
a b c c
ab a b
3 2 3
2 2
⋅
⋅
 = a b c
a b
3 2 4
3 3 = 
a b c
a b
b
/ /
/ /
3 2 4
3 3 = 
c
b
4
 
Practice Problems - Division of Algebraic Fractions 
Section 1.5c Case IV Practice Problems - Divide the following algebraic fractions. Simplify 
the answer to its lowest term. 
1. x y
x y
xy
2
3 2 ÷ = 2. 
uv w
vw
uv
w
2
2
3
÷ = 3. a b c a b
ac
2 2 4
2
2
÷ = 
4. xyz
x z
x z
yz2 3
2 2
÷ = 5. uv
v
u uv
2
3
22
3
÷





 ÷ = 6. 
x y z
xz
x y
yz
2 2
2
3
4
÷ ÷





 = 
 
Mastering Algebra - Advanced Level 1.5d Math Operations Involving Complex Algebraic Fractions 
Hamilton Education Guides 107 
1.5d Math Operations InvolvingComplex Algebraic Fractions 
A simple algebraic fraction is a fraction in which neither the numerator nor the denominator 
contains a fraction with variables. For example, a
5
, a
a
−1 , 2 1
3
x − , 3
2 12
x
x x+ −
, and −
+ +
5
2 12a a
 are 
examples of simple algebraic fractions. A complex algebraic fraction is a fraction in which 
either the numerator or the denominator (or both) contains an algebraic fraction. For example, 
1 1
5
−
w , 
2
5
1
2
x
x
x
−
, 
a a
b
a
b
+
−1
, and x
x
1 1+
 are examples of complex algebraic fractions. 
Note that an easy way to change complex algebraic fractions to simple algebraic fractions is by 
multiplying the outer numerator by the outer denominator and the inner denominator by the inner 
numerator. For example, given the complex fraction 
x
x
2
5
2
9
; first obtain the numerator of the 
simple fraction by multiply x2 , the outer numerator, by 9 , the outer denominator. Next, obtain 
the denominator of the simple fraction by multiply 5 , the inner denominator, by 2x , the inner 
numerator. Therefore, the complex fraction 
x
x
2
5
2
9
 can be written as x
x
2 9
5 2
⋅
⋅
 = 9
10
2x
x
 = 9
10
2x
x
x
/
/
 = 9
10
x 
which is a simple fraction. In this section addition, subtraction, multiplication, and division of 
complex algebraic fractions (Cases I through III) are reviewed. 
Case I Addition and Subtraction of Complex Algebraic Fractions 
Complex algebraic fractions are added or subtracted using the following steps: 
Step 1 Add or subtract the algebraic fractions in both the numerator and the denominator. 
Note that the same process used in simplifying integer (arithmetic) fractions applies to 
algebraic fractions. 
Step 2 Change the complex algebraic fraction to a simple fraction . Reduce the algebraic 
fraction to its lowest term, if possible. 
Examples with Steps 
The following examples show the steps as to how complex algebraic expressions are added and 
subtracted: 
Example 1.5-45 
 
3 1
1
3 2
2
2
x y
xy
x y
xy
−
+
 = 
 
 
Mastering Algebra - Advanced Level 1.5d Math Operations Involving Complex Algebraic Fractions 
Hamilton Education Guides 108 
Solution: 
 Step 1 
3 1
1
3 2
2
2
x y
xy
x y
xy
−
+
 = 
3 1
1
1
1
3 2
2
2
x y
xy
x y
xy
−
+
 = 
( ) ( )
( ) ( )
3 1 1
1
1 1
1
3 2
2 2
2
x y xy
xy
x y xy
xy
⋅ + ⋅
⋅
⋅ + ⋅
⋅
 = 
3 3 2
2 2
2
x y xy
xy
x y xy
xy
+
+
 
 
 Step 2 
3 3 2
2 2
2
x y xy
xy
x y xy
xy
+
+
 = 
( )
( )
3 3 2 2
2 2
x y xy x y
y
xy x y xy
+ ⋅ /
/ / ⋅ +
/
 = 
( )3 3 2
2 2
x y xy y
x y xy
+
+
 = 
( )
( )
/ / +
/ / +
xy x y y
xy x y
3 12
 = 
( )y x y
x y
3 12 +
+
 
Example 1.5-46 
 
1 1
1 1
a b
a b
a
b
+
−
− = 
Solution: 
 Step 1 
1 1
1 1
a b
a b
a
b
+
−
− = 
( ) ( )
( ) ( )
1 1
1 1
⋅ + ⋅
⋅
⋅ − ⋅
⋅
−
b a
a b
b a
a b
a
b
 = 
b a
ab
b a
ab
a
b
+
−
− 
 
 Step 2 
b a
ab
b a
ab
a
b
+
−
− = ( )( )
b a ab
ab b a
a
b
+ ⋅ / /
/ / ⋅ −
− = b a
b a
a
b
+
−
− = 
( )[ ] ( )[ ]
( )
b a b a b a
b a b
+ ⋅ − ⋅ −
− ⋅
 
 
 = ( )
b ab ab a
b b a
2 2+ / / − / / +
−
 = ( )
a b
b b a
2 2+
−
 
Additional Examples - Addition and Subtraction of Complex Algebraic Fractions 
The following examples further illustrate addition and subtraction of complex algebraic fractions: 
Example 1.5-47 
2 2
3
4 2
9
−
−
x
x
 = 
2
1
2
3
4
1
2
9
−
−
x
x
 = 
( ) ( )
( ) ( )
2 3 2 1
1 3
4 9 2 1
1 9
⋅ − ⋅
⋅
⋅ − ⋅
⋅
x
x
x
x
 = 
6 2
3
36 2
9
x
x
x
x
−
− = 
( )
( )
6 2 9
3 36 2
x x
x x
− ⋅
⋅ −
 = ( )( )
18 3 1
6 18 1
x x
x x
−
−
 = ( )( )
/ / / −
/ / −
18
3
3 1
6 18 1
x x
x x
 = 
( )3 3 1
18 1
x
x
−
−
 
Example 1.5-48 
1
8
1
4
+
+
a
a
 = 
1
8 1
1
1
4
+
+
a
a = 
( ) ( )
( ) ( )
1 1 8
8 1
4 1 1
1 4
⋅ + ⋅
⋅
⋅ + ⋅
⋅
a
a
 = 
1 8
8
4 1
4
+
+
a
a = 
( )
( )
1 8 4
8 4 1
+ ⋅
⋅ +
a
a
 = ( )( )
1 8 4
8
2
4 1
+ ⋅ /
/⋅ +
a
a
 = ( )
1 8
2 1 4
+
+
a
a 
 
Mastering Algebra - Advanced Level 1.5d Math Operations Involving Complex Algebraic Fractions 
Hamilton Education Guides 109 
Example 1.5-49 
8
4
6
3
1
3 2
2
3
2
a b
a b
a
a
ab
a
+ = 
( ) ( )
( ) ( )
8 3
4 6
1
1
3 2 2
2 3
a b a
a b a
ab
a
⋅
⋅
+ = / /
/ / /
+
⋅
⋅
/ /
/
24
24 1 1
5 2
5
a b
b
a b
ab a = b a b
1 1
2
+ = b a b+
2
1
 = b a b+ 2 = ( )b a1 2+ 
Example 1.5-50 
x
y
y
x
xy xy
+
+
3 5 = 
( ) ( )x x y y
x y
xy
⋅ + ⋅
⋅
+3 5 = 
x y
xy
xy
2 2
8
+
 = 
( )x y xy
xy
2 2
8
+ ⋅ / /
⋅ / /
 = x y
2 2
8
+ 
Example 1.5-51 
1
2
1
3
2
3
5
2
4x y
x y
−
+
− = 
( ) ( )
( ) ( )
1 3 1 2
2 3
2 2 5 3
3 2
4
⋅ − ⋅
⋅
⋅ + ⋅
⋅
−
y x
x y
y x
x y
 = 
3 2
6
4 15
6
4
y x
xy
y x
xy
−
+
− = 
( )
( )
3 2 6
6 4 15
4
y x xy
xy y x
− ⋅ / / /
/ / / ⋅ +
− = 
3 2
4 15
4y x
y x
−
+
− = 
3 2
4 15
4
1
y x
y x
−
+
− 
 
= 
( ) ( )
( )
3 2 1 4 4 15
4 15 1
y x y x
y x
− ⋅ − ⋅ +
+ ⋅
 = 
3 2 16 60
4 15
y x y x
y x
− − −
+
 = 
( ) ( )3 16 60 2
4 15
y y x x
y x
− + − −
+
 = − −
+
13 62
4 15
y x
y x 
Example 1.5-52 
3 1
1
2
1
3
−
+
+
+
x
x
 = 
3
1
1
1
2
1
3
1
−
+
+
+
x
x
 = 
( )[ ] ( )
( )
( ) ( )[ ]
( )
3 1 1 1
1 1
2 1 3 1
1 1
⋅ + − ⋅
⋅ +
⋅ + ⋅ +
+ ⋅
x
x
x
x
 = 
3 3 1
1
2 3 3
1
x
x
x
x
+ −
+
+ +
+
 = 
3 2
1
3 5
1
x
x
x
x
+
+
+
+
 = ( ) ( )( ) ( )
3 2 1
1 3 5
x x
x x
+ ⋅ / + /
/ + / ⋅ +
 = 3 23 5
x
x
+
+
 
Example 1.5-53 
x
x
x
+
+
+
−
3
2
4
3
1
 = 
x
x
x
+
+
+
−
3
2
1
4
3
1
1
 = 
( ) ( )[ ]
( )
( ) ( )[ ]
( )
x x
x
x
x
⋅ + ⋅ +
+ ⋅
⋅ − ⋅ +
+ ⋅
1 2 3
3 1
4 1 1 3
3 1
 = 
x x
x
x
x
+ +
+
− −
+
2 6
3
4 3
3
 = 
3 6
3
1
3
x
x
x
x
+
+
− +
+
 = ( ) ( )( ) ( )
3 6 3
3 1
x x
x x
+ ⋅ / + /
/ + / ⋅ − +
 = 
( )3 2
1
x
x
+
−
 
Practice Problems - Addition and Subtraction of Complex Algebraic Fractions 
Section 1.5d Case I Practice Problems - Simplify the following complex algebraic fractions. 
Reduce the answer to its lowest term. 
1. 
2 1
5
3 2
15
−
−
a
a
 = 2. 
2
4
2 1
3 2
2
2
x y z
x z
x
xy
− = 3. 
2 1
2 2
3
3
a a
a
+
+
 = 
4. a
a
a
+
1
2
 = 
5. 
x
y
y
x
−
−
3
3
 = 6. 
1 1
4
3
4
1
x x
x
−
+
+
+
 = 
 
Mastering Algebra - Advanced Level 1.5d Math Operations Involving Complex Algebraic Fractions 
Hamilton Education Guides 110 
Case II Multiplication of Complex Algebraic Fractions 
Complex algebraic fractions are multiplied by one another using the following steps: 
Step 1 Multiply the numerator and the denominator of the algebraic fraction terms by one 
another. 
Step 2 Change the complex algebraic fraction to a simple fraction. Reduce the algebraic 
fraction to its lowest term, if possible. 
Examples with Steps 
The following examples show the steps as to how complex algebraic fractions are multiplied: 
Example 1.5-54 
 
x y z
z
z
x y
xy
2 2
2 3 3
1
⋅
 = 
Solution: 
 Step 1 
x y z
z
z
x y
xy
2 2
2 3 3
1
⋅
 = 
x y z z
z x y
xy
2 2
2 3 3
1
⋅
⋅ = 
x y z
x y z
xy
2 2 2
3 3 2
1 
 
 Step 2 
x y z
x y z
xy
2 2 2
3 3 2
1 = 
x y z xy
x y z
2 2 2
3 3 2 1
⋅
⋅
 = x y z
x y z
3 3 2
3 3 2 = 
x y z
x y z
/ / /
/ / /
3 3 2
3 3 2 = 
1
1
 = 1 
Example 1.5-55 
 
u v
w
w
v
uw
2 2 3
4
2
⋅
 = 
Solution: 
 Step 1 
u v
w
w
v
uw
2 2 3
4
2
⋅
 = 
u v
w
w
v
uw
2 2 3
4
2
1
⋅
 = 
u v w
w v
uw
2 2 3
4
2
1
⋅
⋅ = 
u v w
wv
uw
2 2 3
4
2
1
 
 
 Step 2 
u v w
wv
uw
2 2 3
4
2
1
 = u v w
wv uw
2 2 3
4 2
1⋅
⋅
 = u v w
uv w
2 2 3
4 2 = 
u
u
v w
u v
v
w
/ / /
/ //
2 2 3
4
2
3 = 
u
v2
 
 
Mastering Algebra - Advanced Level 1.5d Math Operations Involving Complex Algebraic Fractions 
Hamilton Education Guides 111 
Additional Examples - Multiplication of Complex Algebraic Fractions 
The following examples further illustrate how to multiply complex algebraic fractions: 
Example 1.5-56 
xyz x
y z
x
⋅
2
2 2
2 = 
xyz x
y z
x
1
22 2
2
⋅
 = 
xyz x
y z
x
⋅
⋅
2
1 2 2
2 = 
2 2
2 2
2
x yz
y z
x
 = 
2
1
2
2 2
2
x yz
y z
x
 = 2 1
2
2 2 2
x yz
y z x
⋅
⋅
 = 2
2
2 2 2
x yz
x y z
 = 2
2
2 2 2
x yz
x y
y
z
z
/
/ / /
/ / = 2yz 
Example 1.5-57 
2
3
62 2
3 3
x y
x
x
x y
⋅
 = 2
6
3
2 2
3 3
x y x
x x y
⋅
⋅
 = 2
6
3
3 2
4 3
x y
x y
 = 
2
1
6
3
3 2
4 3
x y
x y
 = 2 3
1 6
4 3
3 2
⋅
⋅
x y
x y
 = 6
6
4 3
3 2
x y
x y
 = /
/
/ /
/ /
6
6
4 3
3 2
x
x
y
y
x y
 = xy
1
 = xy 
Example 1.5-58 
ab
a b a
ab
3 2
3
3
⋅
 = ab
a b a
ab
3 2
31
3
⋅
 = ab
a b a
ab
3 2
3
3
1
⋅
⋅
 = ab
a b
ab
3 4 2
3
 = 
ab
a b
ab
1
3 4 2
3
 = ab ab
a b
⋅
⋅
3
4 21 3
 = a b
a b
2 4
4 23
 = a b
b
a
a
b
/ /
/ /
2 4
2
4
2
23
 = b
a
2
23
 
Example 1.5-59 
uv
u v
u
v
u
v
⋅ ⋅
1 2
2 2
3
4
3
 = 
uv
u v
u
v
u
v
1
1 2
2 2
3
4
3
⋅ ⋅
 = 
uv u
u v v
u
v
⋅ ⋅
⋅ ⋅
1 2
1
3
2 2 4
3
 = 
2 4
2 6
3
u v
u v
u
v
 = 2
4 3
2 6
u v v
u v u
⋅
⋅
 = 2
4 4
3 6
u v
u v
 = 2
4 4
3 6
2
u
u
v
u v
v
/ /
/ / = 
2
2
u
v
 
Example 1.5-60 
xyz
xy
x
y
x
y
y
x
2 2
3
2
3
⋅
⋅
 = 
xyz x
xy y
x y
y x
2 2
3
2
3
⋅
⋅
⋅
⋅
 = 
x yz
xy
xy
x y
3 2
4
2
3
 = x yz x y
xy xy
3 2 3
4 2
⋅
⋅
 = x y z
x y
6 2 2
2 6 = 
x
x
y z
x y
y
/ /
/ /
6
4
2 2
2 6
4
 = x z
y
4 2
4 
Practice Problems - Multiplication of Complex Algebraic Fractions 
Section 1.5d Case II Practice Problems - Multiply the following complex algebraic 
expressions. Simplify the answer to its lowest term. 
1. y
x y
xy
x y
x
2
2 2 3
4
2
⋅
 = 
2. 
1
2
4
1
3 2
2
x y
x
y
xy
⋅
 3. 
x
x
x x
x
2
3
2
52
3
18
⋅
 
4. 
x yz x
y
x
x z
y z
2 2
2 3 4
2 3
2 3
1
⋅





 ⋅
 5. 
2 4 1
3
8
3
2 2
3
3
4
a b
a
ab
ab
⋅





 ⋅
 6. 
u v w
w w
uv
3 3 2
3
2
1
3
1
6
⋅
 
 
Mastering Algebra - Advanced Level 1.5d Math Operations Involving Complex Algebraic Fractions 
Hamilton Education Guides 112 
Case III Division of Complex Algebraic Fractions 
Complex algebraic fractions are divided by one another using the following steps: 
Step 1 Write the algebraic expression in fraction form, i.e., write x y2 3 as x y
2 3
1
. 
Step 2 a. Invert the second fraction in either the numerator or the denominator, or both. 
 b. Change the division sign to a multiplication sign. 
 c. Multiply the numerator and the denominator of the algebraic fraction terms by one 
 another. 
Step 3 Change the complex algebraic fraction to a simple fraction. Reduce the algebraic 
fraction to its lowest term, if possible. 
Examples with Steps 
The following examples show the steps as to how complex algebraic fractions are divided: 
Example 1.5-61 
 
a b c
b c
a
ab
a
b
3 2
2 3
3
3
2
÷
÷
 = 
Solution: 
 Step 1 
a b c
b c
a
ab
a
b
3 2
2 3
3
3
2
÷
÷
 = 
a b c
b c
a
ab
a
b
3 2
2 3
3
3
2
1
1
÷
÷
 
 
 Step 2 
a b c
b c
a
ab
a
b
3 2
2 3
3
3
2
1
1
÷
÷
 = 
a b c
b c a
ab
a b
3 2
2 3 3
3 2
1
1
⋅
⋅
 = 
a b c
b c a
ab
a b
3 2
2 3 3
3 2
1
1
⋅
⋅
⋅
⋅
 = 
a b c
a b c
ab
a b
3 2
3 2 3
3 2
 
 
 Step 3 
a b c
a b c
ab
a b
3 2
3 2 3
3 2
 = a b c a b
a b c ab
3 2 3 2
3 2 3
⋅
⋅
 = a b c
a b c
6 4
4 3 3 = 
a
a
b
b
c
a b c
c
/ /
/ / /
/6
2
4
4 3 3
2
 = a b
c
2
2 
Example 1.5-62 
 3
2
2
2
xyz
xy
z
z
xy
xyz
÷
÷
 = 
 
 
Mastering Algebra - Advanced Level 1.5d Math Operations Involving Complex Algebraic Fractions 
Hamilton Education Guides 113 
Solution: 
 Step 1 3
2
2
2
xyz
xy
z
z
xy
xyz
÷
÷
 = 3
1
1
2
2
2
xyz
xy
z
z
xy
xyz
÷
÷
 
 
 Step 2 3
1
1
2
2
2
xyz
xy
z
z
xy
xyz
÷
÷
 = 3
1
1
2
2
2
xyz
xy
z z
xy
xyz
÷
⋅
 = 3
1
1
2
2
2
xyz
xy
z z
xy
xyz
÷
⋅
⋅ = 3
1
3
2
2
xyz
xy
z
xy
xyz
÷ 
 
 Step 3 3
1
3
2
2
xyz
xy
z
xy
xyz
÷ = 3
1
2
3 2
xyz xy xyz
z xy
÷
⋅
⋅
 = 3
1
2 2 2
2 3
xyz x y z
xy z
÷ = 3
1
2 3
2 2 2
xyz xy z
x y z
⋅ 
 
 = 3
1
2 3
2 2 2
xyz xy z
x y z
⋅
⋅
 = 3
2 3 4
2 2 2
x y z
x y z
 = 3
2 3 4
2
2 2 2
x y
y
z
z
x y z
/ / /
/ / / = 
3
1
2yz = 3 2yz 
 
Additional Examples - Division of Complex Algebraic Fractions 
The following examples further illustrate how to divide complex algebraic fractions: 
Example 1.5-63 
xyz x z
x
xy
2
2 3
3
÷
 = 
xyz x z
x
xy
2 2 3
1 3
÷
 = 
xyz x
x z
xy
2
2 31
3
⋅
 = 
xyz x
x z
xy
2
2 3
3
1
⋅
⋅ = 
3 2 2
2 3
x yz
x z
xy
 = 
3
1
2 2
2 3
x yz
x z
xy = 
3 12 2
2 3
x yz
x z xy
⋅
⋅
 
 
= 3
2 2
3 3
x yz
x yz
 = 3
2 2
3 3
x yz
x
x
y z
z
/ /
/ /
/
/
 = 3xz 
Example 1.5-64 
2
2 4
2 2 2
3
x
x y
x
x y
x
÷
 = 
2
2
42 3
2 2
x
x y
x
x
x y
⋅
 = 
2
4
2
2 3
2 2
x
x y x
x x y
⋅
⋅
 = 
2
4
2
5
3 2
x
x y
x y
 = 2 2
4
3 2
5
⋅
⋅
x y
x x y
 = 4
4
3 2
6
x y
x y
 = /
/ /
/ /
/
4
4
3 2
6
3
x y
y
x
x
y
 = y
x3
 
Example 1.5-65 
xyz
x
x y
x
x
y
2
2 2
3÷
 = 
xyz
x
x
x y
x
y
2
3
2 2⋅
 = 
xyz x
x x y
x
y
⋅
⋅
3
2 22 = 
x yz
x y
x
y
4
3 22 = x yz y
x y x
4
3 22
⋅
⋅
 = x y z
x y
4 2
4 22
 = x y z
x y
/ /
/ /
4 2
4 22
 = z2 
 
Mastering Algebra - Advanced Level 1.5d Math Operations Involving Complex Algebraic Fractions 
Hamilton Education Guides 114 
Example 1.5-66 
a b
a
b
a b
b
a
b
2 2 3
2 3
2
÷
÷
 = a b
a
b
a b
b
b
a
2 2 3
2 2
3
÷
⋅
 = a b
a
b
a b b
b a
2 2 3
2 2
3
÷
⋅
⋅
 = a b
a
b
a b
a b
2 2 3
2 3
3
÷ = a b a a b
b a b
2 2
3
3 2 3÷
⋅
⋅
 
 
= a b a b
a b
2 2
4
2 6÷ = a b
a b
a b
2 2
2 6
4⋅ = 
a b a b
a b
2 2 2 6
41
⋅ = a b a b
a b
2 2 2 6
41
⋅
⋅
 = a b
a b
4 8
4 = 
a b
b
a b
/ /
/ /
4 8
7
4 = 
b7
1
 = b7 
Example 1.5-67 
1
3
2 3
2 2a bc
b c
a
a b
b
÷ = 
1
1 1
3
2 3
2 2a bc
b c
a
a b
b
÷ = 1
1 1
2 3
3 2 2
⋅
⋅
÷
⋅
⋅
b c
a bc
a b
a b
 = b c
a bc
ab
a b
2 3
3 2 2÷ = 
b c
a bc
a b
ab
2 3
3
2 2
⋅ = b c a b
a bc ab
2 3 2 2
3
⋅
⋅
 
 
= a b c
a b c
2 4 3
4 2 = 
a b
b
c
c
a
a
b c
/ / /
/ / /
2 4
2
3
2
4
2
2 = 
b c
a
2 2
2 
Example 1.5-68 
u v
uw
uv
w
uw
u
uv
w
2 2
2
2
2
÷
÷
 = 
u v
uw
w
uv
uw
u
w
uv
2 2 2
2
2
⋅
⋅
 = 
u v w
uw uv
uw w
u uv
2 2 2
2
2
⋅
⋅
⋅
⋅
 = 
u v w
u vw
uw
u v
2 2 2
2
2
3
2
 = u v w u v
u vw uw
2 2 2 3
2 22
⋅
⋅
 = u v w
u vw
5 3 2
3 32
 = u
u
v
v
w
u v w
w
/ / /
/ //
5
2
3
2
2
3 32
 = u vw
2 2
2 
Example 1.5-69 
3
2
2
2x
x
y
x y
÷
÷
 = 
3
1
2 1
2
2
x x
y
x y
÷
÷
 = 
3
1
2
1
2
2
x y
x
x
y
⋅
⋅
 = 
3
1
1
2
2
2
x y
x
x
y
⋅
⋅
⋅
⋅
 = 
3
2
2
2
xy
x
x
y
 = 3 2
2
2
xy y
x x
⋅
⋅
 = 6
3
3
xy
x
 = 6
3
3
2
/
/
xy
x
x
 = 6
3
2
y
x
 
Practice Problems - Division of Complex Algebraic Fractions 
Section 1.5d Case III Practice Problems - Divide the following complex algebraic fractions. 
Simplify the answer to its lowest term. 
1. 
xy
x
x
xy
x
3
2
2
2
÷
 = 2. 
1 1
2
a b
a
ab
b
÷
÷
 = 3. xy xy
x
x
÷
÷











1
3 = 
4. 
u vw
w
u v
w
w
2
2
3
3
÷
 = 5. 
ab a b
b
a a
b
3
2 2
÷
÷
 = 6. 
z
z
z
z z
3
2
3
2
2
2
÷
÷
 = 
 
Mastering Algebra - Advanced Level Quick Reference to Chapter 2 Problems 
Hamilton Education Guides 115 
Chapter 2 
Functions of Real and Complex Variables 
 
Quick Reference to Chapter 2 Problems 
2.1 Introduction to Functions of Real Variables ........................................................... 117 
( )2+af = ; ( ) ( )afhaf −+ = ; ( ) 42xf = 
2.2 Math Operations Involving Functions of Real Variables ...................................... 124 
( ) ( )1213 323 −+++ xxx = ; 13
12
1 23
3
+++
−
xx
x
 = ; 
13
1
13
3
2323 ++
+
−
++ xx
a
xx
 = 
2.2.1 Determining Odd and Even Functions 127 
( ) ( ) oddisfunctionthexfxfif −=− ; ( ) ( ) evenisfunctionthexfxfif =− 
2.3 Composite Functions of Real Variables ...................................................................129 
Case I – Computing Composite Functions Using ( )xf and ( )xg 129 
Find ( )( )xgfgf = given ( ) ( ) 21 xxgandxxf =+= 
Case II – Computing Composite Functions Using ( )xf , ( )xg ,and ( )xh 134 
Find ( )( )( )xhgfhgf = given ( ) ( ) ( ) 2,1,2 xxhandxxgxxf =+== 
2.4 One-to-One and Inverse Functions of Real Variables ............................................ 139 
2.4.1 One-to-One Functions 139 
Case I – The Function is Represented by a Set of Ordered Pairs 139 
( ) ( ) ( )[ ]9,7,6,2,4,1=f = ; ( ) ( ) ( ) ( )[ ]9,6,7,3,6,2,4,2=f = 
Case II – The Function is Represented by an Equation 139 
( ) 13 += xxf = ; ( ) 210 xxf += = ; ( ) 1+= xxf = 
2.4.2 Inverse Functions 142 
Case I – The Function is Represented by a Set of Ordered Pairs 142 
( ) ( ) ( )[ ]6,8,5,2,3,1=f = ; ( ) ( ) ( ) ( )[ ]10,6,8,3,8,1,5,1=f = 
Case II – The Function is Represented by an Equation 143 
( ) 12 += xxf = ; ( ) 22 −= xxf = ; ( ) 31 xxf += = 
 
Mastering Algebra - Advanced Level Quick Reference to Chapter 2 Problems 
Hamilton Education Guides 116 
2.5 Complex Numbers and Functions of Complex Variables ...................................... 151 
169 −+− = ; ( )ig 31− = ; ( )( )igf 23 + = 
2.6 Math Operations Involving Complex Numbers ...................................................... 159 
Case I – Addition and Subtraction of Complex Numbers 159 
( ) ( )ii 4523 −++ = ; ( ) ( )93452 −−−−+ = ; ( )[ ] ( )iii 3513272 −−++ = 
Case II – Multiplication of Complex Numbers 160 
( )( )ii +−−− 532 = ; ( )( )ii 2553 −+ = ; ( )[ ] ( )iii −+− 255 3 = 
Case III – Division of Complex Numbers 161 
i
i
23
52
+
− = ; 
i
i
253
23
−
+− = ; 383
51
i
i
−
− = 
Case IV – Mixed Operations Involving Complex Numbers 163 
i
i
i
i
+
+
×
+
+
3
2
1
32 = ; 
i
i
i
i
−
+
÷
+
+
1
1
31
2 = ; 
i
i
i 23
32
1
4
−
+
+
+
 = 
 
 
Hamilton Education Guides 117 
Chapter 2 – Functions of Real and Complex Variables 
 
The objective of this chapter is to introduce students to operations involving functions of real and 
complex variables and operations involving complex numbers. Relations and functions are 
introduced in Sections 2.1. Addition, subtraction, multiplication, and division of functions are 
addressed in Section 2.2. Composition of two or more functions and its computation is discussed 
in Section 2.3. Identification of a one-to-one function as well as computation of inverse 
functions are addressed in Section 2.4. Complex numbers and functions of complex variables 
are introduced in Section 2.5. Finally, math operations involving addition, subtraction, 
multiplication, and division of complex numbers are addressed in Section 2.6. Each section is 
concluded by solving examples with practice problems to further enhance the student ability. 
2.1 Introduction to Functions of Real Variables 
Recall that real numbers are the numbers on the real number line from ∞− to ∞+ . These are the 
“normal” every day numbers that we use to count using addition, subtraction, multiplication, and 
division methods. We identify these numbers as real because, as we will see later in this chapter, 
the need will arise to define a different set of numbers called imaginary numbers with some 
unique set of properties to perform math operations. Note that the functions we are considering 
in this section have real numbers as variables. 
In order to proceed with the definition of a function we first need to learn a few terms. An 
ordered pair is defined by two numbers each called a component. The first pair, usually 
denoted by x , is referred to as the first component and the second pair, usually denoted by y , is 
referred to as the second component. In this section only real number components are 
addressed. The domain is defined as the set of all the first components of the ordered pairs. The 
range is defined as the set of all the second components of the ordered pairs. For example, in the 
relation defined by ( ) ( ) ( ) ( ) ( ){ }9,5,8,3,6,1,3,2,2,1 the domain is { }5,3,2,1 and and the range is 
{ }9,8,6,3,2 and . Note that ( )xf , which is the same as y , represents an element in the range of a 
function. For example, in the equation ( ) 122 ++== xxyxf since the y value depends on the 
value of x , we call y the dependent variable and x the independent variable. Having 
addressed these terms we can now proceed with defining a function. 
In general, a function is denoted by symbols such as f , g , h , p , q , r , s , and t . A symbol such 
as ( )xf , read as “ xoff ” represents the range value associated with the domain which is the x 
value. To understand functions we first need to know about relations. Relations are often 
defined by equations in two variables. In a relation each element of the domain is paired with 
more than one element of the range. For example: 1. The ordered pairs ( ) ( ) ( ) ( ){ }7,3,6,2,4,1,3,1 
belong to a relation because each element of the domain is not paired with only one element of 
the range, i.e., two ordered pairs, ( ) ( )4,1,3,1 , have the same first components or, 2. The equation 
y = 1+± x defines a relation. This is because for each value of 0x there are two values of y , 
i.e., at 0=x the variable 1±=y or at 2=x the variable 3±=y , etc. 
On the other hand, a function is defined as a relation in which each element of the domain is 
paired with only one element of the range. For example: 1. The ordered pairs 
( ) ( ) ( ) ( ){ }6,4,5,3,3,2,2,1 belong to a function because each element of the domain is paired with 
only one element of the range or, 2. The equation y = 3+x defines a function. This is because 
Mastering Algebra - Advanced Level 2.1 Introduction to Functions of Real Variables 
Hamilton Education Guides 118 
for each value of x there is only one value of y , i.e., at 0=x the variable 3=y or at 2=x the 
variable 5=y , etc. 
Example 2.1-1: Given the following equations, find the corresponding y values (range) for the 
given x values (domain). 
a. 032 =+ yx at 10,2,1,0 ==−== xandxxx b. 42 += xy at 5,3,0,1 ===−= xandxxx 
c. 12 += xy at 10,4,2,0 ==== xandxxx d. 62 =− yx at 2,1,3,0 =−=−== xandxxx 
e. 92 += xy at 3,1,4,0 =−=−== xandxxx 
Solutions: 
a. Given 032 =+ yx , at 0=x the y value is equal to ( ) 0302 =+× y ; 030 =+ y ; 03 =y ; 0=y 
 
at 1−=x the y value is equal to ( ) 0312 =+−× y ; 032 =+− y ; 23 =y ; 
3
2
=y ; 66.0=y 
 
at 2=x the y value is equal to ( ) 0322 =+× y ; 034 =+ y ; 43 −=y ; 
3
4
−=y ; 33.1−=y 
 
at 10=x the y value is equal to ( ) 03102 =+× y ; 0320 =+ y ; 203 −=y ; 
3
20
−=y ; 66.6−=y 
 
Therefore, the ordered pairs are ( ) ( ) ( ) ( ){ }66.6,10,33.1,2,66.0,1,0,0 −−− 
b. Given 42 += xy , at 1−=x the y value is equal to ( ) 41 2 +−=y ; 41+=y ; 5=y 
 
at 0=x the y value is equal to 402 +=y ; 40+=y ; 4=y 
 
at 3=x the y value is equal to 432 +=y ; 49+=y ; 13=y 
 
at 5=x the y value is equal to 452 +=y ; 425+=y ; 29=y 
 
Therefore, the ordered pairs are ( ) ( ) ( ) ( ){ }29,5,13,3,4,0,5,1− 
c. Given 12 += xy , at 0=x the y value is equal to ( ) 102 +×=y ; 10+=y ; 1=y ; 1=y 
 
at 2=x the y value is equal to ( ) 122 +×=y ; 14+=y ; 5=y ; 24.2=y 
 
at 4=x the y value is equal to ( ) 142 +×=y ; 18+=y ; 9=y ; 3=y 
 
at 10=x the y value is equal to ( ) 1102 +×=y ; 120+=y ; 21=y ; 58.4=y 
 
Therefore, the ordered pairs are ( ) ( ) ( ) ( ){ }58.4,10,3,4,24.2,2,1,0 
d. Given 62 =− yx , at 0=x the y value is equal to 620 =− y ; 62 =− y ; 
2
6
−=y ; 3−=y 
 
at 3−=x the y value is equal to 623 =−− y ; 362 +=− y ; 92 =− y ; 
2
9
−=y ; 5.4−=y 
Mastering Algebra - Advanced Level 2.1 Introduction to Functions of Real Variables 
Hamilton Education Guides 119 
at 1−=x the y value is equal to 621 =−− y ; 162 +=− y ; 72 =− y ; 
2
7−=y ; 5.3−=y 
 
at 2=x the y value is equal to 622 =− y ; 262 −=− y ; 42 =− y ; 
2
4
−=y ; 2−=y 
 
Therefore, the ordered pairs are ( ) ( ) ( ) ( ){ }2,2,5.3,1,5.4,3,3,0 −−−−−− 
e. Given 92 += xy , at 0=x the y value is equal to 902 +=y ; 90+=y ; 9=y 
 
at 4−=x the y value is equal to ( ) 94 2 +−=y ; 916+=y ; 25=y 
 
at 1−=x the y value is equal to ( ) 91 2 +−=y ; 91+=y ; 10=y 
 
at 3=x the y value is equal to 932 +=y ; 99+=y ; 18=y 
 
Therefore, the ordered pairs are ( ) ( ) ( ) ( ){ }18,3,10,1,25,4,9,0 −− 
Example 2.1-2: 1. Specify the domain and the range for each of the following ordered pairs. 
2. State which sets constitute a relation or a function. 
a. ( ) ( ) ( ) ( ) ( ){ }11,8,10,6,5,3,3,2,2,1 b. ( ) ( ) ( ) ( ) ( ){ }12,10,8,6,7,4,5,4,2,2 − 
c. ( ) ( ) ( ) ( ){ }8,5,4,2,3,0,2,1− d. ( ) ( ) ( ) ( ) ( ){ }5,2,8,4,6,3,4,2,0,0 
e. ( ) ( ) ( ) ( ){ } ( )14,10,12,8,7,4,6,2,3,2 −− f. ( ) ( ) ( ) ( ) ( ){ }15,9,12,7,6,5,4,3,1,2 
Solutions: 
a. The domain and the range for the ordered pairs ( ) ( ) ( ) ( ) ( ){ }11,8,10,6,5,3,3,2,2,1 are: { }8,6,3,2,1 and 
{ }11,10,5,3,2 . Since each domain value corresponds with only one range value it is a function. 
b. The domain and the range for the ordered pairs ( ) ( ) ( ) ( ) ( ){ }12,10,8,6,7,4,5,4,2,2 − are: { }10,6,4,2 
and { }12,8,7,5,2− . Since each domain value does not corresponds with only one range value it 
is a relation. 
c. The domain and the range values for the ordered pairs ( ) ( ) ( ) ( ){ }8,5,4,2,3,0,2,1− are: { }5,2,0,1− 
and { }8,4,3,2 . Since each domain value corresponds with only one range value it is a function. 
d. The domain and the range values for the ordered pairs ( ) ( ) ( ) ( ) ( ){ }5,2,8,4,6,3,4,2,0,0 are: 
{ }4,3,2,0 and { }8,6,5,4,0 . Since each domain value does not correspond with only one range 
value it is a relation. 
e. The domain and the range values for the ordered pairs ( ) ( ) ( ) ( ){ } ( )14,10,12,8,7,4,6,2,3,2 −− are: 
{ }10,8,4,2− and { }14,12,7,6,3 . Since each domain value does not correspond with only one 
range value it is a relation. 
f. The domain and the range values for the ordered pairs ( ) ( ) ( ) ( ) ( ){ }15,9,12,7,6,5,4,3,1,2 are: 
{ }9,7,5,3,2 and { }15,12,6,4,1 . Since each domain value corresponds with only one range 
value it is a function. 
Mastering Algebra - Advanced Level 2.1 Introduction to Functions of Real Variables 
Hamilton Education Guides 120 
Example 2.1-3: State which of the following equations define a function. 
a. 2=− yx b. 252 =+ yx c. 2522 =+ yx d. 2xy = 
e. 32 xy = f. 2xy ±= g. 72 −= xy h. 62 += xy 
Solutions: 
a. The equation 2=− yx ; xy −=− 2 ; 2−= xy defines a function because there is only one value 
of y associated with each value of x . 
b. The equation 252 =+ yx ; 225 xy −= defines a function because there is only one value of y 
associated with each value of x . 
c. The equation 2522 =+ yx ; 22 25 xy −= ; 225 xy −±= defines a relation because for each 
value of x there are two values of y . 
d. The equation 2xy = defines a function because there is only one value of y associated with 
each value of x . 
e. The equation 32 xy = ; 3xy ±= defines a relation because for each value of x there are two 
values of y . 
f. The equation 2xy ±= ; xy ±= defines a relation because for each value of x there are two 
values of y . 
g. The equation 72 −= xy defines a function because there is only one value of y associated 
with each value of x . 
h. The equation 62 += xy defines a function because there is only one value of y associated 
with 
each value of x . 
Example 2.1-4: Find the corresponding range values for each of the following functions. Write 
the ordered pair for each case. 
a. ( ) xxf −= for ( )1−f , ( )0f , ( )2f , ( )2+af , and ( )af − 
b. ( ) 132 −+= xxxf for ( )1−f , ( )0f , ( )5f , ( )af − , and ( )1+af 
c. ( ) 13 += xxf for ( )0f , ( )1f , ( )3f , ( )2af , and ( )af 
d. ( )
1
1
+
=
x
xf for ( )2−f , ( )0f , ( )3f , ( )1+af , and ( )af − 
Solutions: 
a. Given ( ) xxf −= , then ( )1−f = ( )1−− = 1 ( )0f = 0 ( )2f = 2− 
 
 ( )2+af = ( )2+− a ( )af − = ( )a−− = a . 
Therefore, the ordered pairs are equal to ( ) ( ) ( ) ( ) ( ){ }aaaa ,,2,2,2,2,0,0,1,1 −−−+−− 
b. Given ( ) 132 −+= xxxf , then ( )1−f = ( ) ( ) 1131 2 −−×+− = 131 −− = 3− 
 
Mastering Algebra - Advanced Level 2.1 Introduction to Functions of Real Variables 
Hamilton Education Guides 121 
( )0f = ( ) 10302 −×+ = 100 −+ = 1− ( )5f = ( ) 15352 −×+ = 11525 −+ = 39 
 
( )af − = ( ) ( ) 132 −−×+− aa = 132 −− aa ( )1+af = ( ) ( ) 1131 2 −+++ aa = 352 ++ aa 
Therefore, the ordered pairs are equal to ( ) ( ) ( ) ( ) ( ){ }35,1,13,,39,5,1,0,3,1 22 +++−−−−−− aaaaaa 
c. Given ( ) 13 += xxf , then ( )0f = ( ) 103 +× = 10+ = 1 = 1 
 
( )1f = ( ) 113 +× = 13+ = 4 = 2 ( )3f = ( ) 133 +× = 19+ = 10 
 
( )2af = ( ) 13 2 +×a = 13 2 +a ( )af = ( ) 13 +×a = 13 +a 
Therefore, the ordered pairs are equal to ( ) ( ) ( ) ( )





 +




 + 13,,13,,10,3,2,1,1,0 22 aaaa 
d. Given ( )
1
1
+
=
x
xf , then ( )2−f = 
12
1
+−
 = 
1
1
− = 1− 
 
( )0f = 
10
1
+
 = 
1
1 = 1 ( )3f = 
13
1
+
 = 
4
1 
 
( )1+af = ( ) 11
1
++a
 = 
2
1
+a
 ( )af − = 
1
1
+− a
 = 
a−1
1 
Therefore, the ordered pairs are equal to ( ) ( )












−
−





+
+




−−
a
a
a
a
1
1,,
2
1,1,
4
1,3,1,0,1,2 
In some instances, for a given function, it is often useful to calculate the change in the range 
value, usually shown by the symbol ∆ , due to a given change, h , in the domain value, i.e., as the 
domain value changes from a to ha + , the corresponding range value also changes from ( )af to 
( )haf + . To calculate this change we use the following equation: 
∆ = ( ) ( )afhaf −+ 
Example 2.1-5: Given the following functions, find ( ) ( )afhaf −+ . 
a. ( ) 23 += xxf b. ( ) 122 −+= xxxf c. ( ) ( )12 += xxxf 
Solutions: 
a. Given ( ) 23 += xxf then ( ) ( )afhaf −+ = ( )[ ] [ ]2323 +−++ aha = 23233 −−++ aha = h3 
 
b. Given ( ) 122 −+= xxxf then ( ) ( )afhaf −+ = ( ) ( )[ ] [ ]1212 22 −+−−+++ aahaha = aahha 2222 +++ 
 
+ 1212 2 +−−− aah = hahh 222 ++ = ( )22 ++ ahh 
 
Mastering Algebra - Advanced Level 2.1 Introduction to Functions of Real Variables 
Hamilton Education Guides 122 
c. Given ( ) ( ) xxxxxf +=+= 2212 then ( ) ( )afhaf −+ = ( ) ( )[ ] ( )aahaha +−+++ 22 22 = ( )ahha 22 22 ++ 
 
+ aaha −−+ 22 = aahaahha −−++++ 222 2422 = hahh ++ 42 2 = ( )142 ++ ahh 
Finally, the notation ( ) baxf = ( ) ( )afbf − is used in some applications in calculus where the 
difference between two specific range values are of interest. 
Example 2.1-6: Given the following functions, find ( ) abxf . 
a. ( ) 13 −= xxf where 2=a and 4=b b. ( ) 322 −+= xxxf where 0=a and 3=b 
c. ( ) 23 3 += xxf where 2−=a and 2=b d. ( ) x
x
xf 21 += where 3=a and 5=b 
Solutions: 
a. Given ( ) 13 −= xxf then ( ) 42xf = ( )[ ] ( )[ ]123143 −⋅−−⋅ = ( ) ( )16112 −−− = 511− = 6 
 
b. Given ( ) 322 −+= xxxf then ( ) 30xf = ( )[ ] ( )[ ]30203323 22 −⋅+−−⋅+ = ( ) 3369 +−+ = 15 
 
c. Given ( ) 23 3 += xxf then ( ) 22−xf = [ ] ( )[ ]223223 33 +−⋅−+⋅ = ( ) ( )224224 +−−+ = 2226+ = 48 
 
d. Given ( ) x
x
xf 21 += then ( ) 53xf = 



 ⋅+−




 ⋅+ 32
3
152
5
1 = ( ) ( )633.0102.0 +−+ = 33.62.10 − = 873. 
In the next section we will address: a. The basic math operations involving functions and b. 
learn how to identify odd and even functions. 
Section 2.1 Practice Problems – Introduction to Functions of Real Variables 
1. Find the corresponding y values. 
 a. 04 =− yx at 3,1,0 =−== xandxx b. 012 =+− xy at 3,3,1 −==−= xandxxc. 012 =+− xy at 5,2,2 −=−== xandxx d. 54 −=+ yx at 4,2,0 =−== xandxx 
 e. 62 2 −= xy at 3,2,0 −=−== xandxx 
2. a. Specify the domain and the range for each of the following ordered pairs. b. State which 
 set constitute a relation or a function. 
 a. ( ) ( ) ( ) ( ) ( ){ }12,8,9,6,6,3,5,2,4,1 b. ( ) ( ) ( ) ( ) ( ){ }11,10,9,7,6,4,3,2,1,1 − 
 c. ( ) ( ) ( ) ( ) ( ){ }12,9,8,6,5,2,2,1,2,1 − d. ( ) ( ) ( ) ( ) ( ){ }8,5,7,2,5,2,6,1,0,0 
 e. ( ) ( ) ( ) ( ) ( ){ }12,10,10,8,5,2,6,1,3,1 −− f. ( ) ( ) ( ) ( ) ( ){ }13,8,10,7,6,5,5,2,3,1 
3. State which of the following equations defines a function. 
 a. 12=+ yx b. 8122 =+ yx c. 22 15 xy −= 
Mastering Algebra - Advanced Level 2.1 Introduction to Functions of Real Variables 
Hamilton Education Guides 123 
 d. 52 xy = e. 92 += xy f. 1062 =− xy 
 g. 42 −= xy h. 84 =− yx i. 92 =+ yx 
4. Find the corresponding range values for each of the following functions. 
 a. ( ) xxxf 22 +−= for ( )4−f , ( )1−f , ( )0f , and ( )1f 
 b. ( ) 12 23 +−= xxxf for ( )2−f , ( )0f , ( )2f , and ( )af − 
 c. ( ) 12 += xxf for ( )2−f , ( )1−f , ( )0f , and ( )2f 
 d. ( )
2
1
2 +
=
x
xf for ( )3−f , ( )1−f , ( )0f , and ( )3f 
5. Given the following functions, find ( ) ( )afhaf −+ . 
 a. ( ) 12 −= xxf b. ( ) 32 2 −= xxf c. ( )
x
axxf
22 −
= d. ( ) ( )( )13 +−= xxxf 
Mastering Algebra - Advanced Level 2.2 Math Operations Involving Functions of Real Variables 
Hamilton Education Guides 124 
2.2 Math Operations Involving Functions of Real Variables 
Two functions ( )f x and ( )xg are added, subtracted, multiplied, and divided using the following 
general operational rules: 
( ) ( )xgxf + = ( )( )xgf + 
( ) ( )xgxf − = ( )( )xgf − 
( ) ( )xgxf ⋅ = ( )( )xgf ⋅ 
( )
( )xg
xf = ( )x
g
f





 where ( ) 0≠xg 
In addition, note that ( )xg
1 = ( )x
g 





 1 , ( )xfα = ( ) ( )xfα , and ( ) ( )xfxf βα ± = ( ) ( )xff βα ± where α 
and β are real numbers. The following examples show basic math operations involving functions. 
Example 2.2-1: Given ( )xf = 13 23 ++ xx and ( )xg = 12 3 −x , find 
a. ( ) ( )xgxf + = b. ( ) ( )xgxf − = c. ( )( ) 3
2
+xg
xf = d. ( ) ( )xgxf ⋅2 = 
e. ( )( ) ( )( )xgxf 25 − = f. ( ) ( )xfxg +
1 = g. ( )( )xgf 32 − = h. ( ) 42 −





x
g
f = 
i. ( ) ( )xgxf 5+ = j. ( )( )xgf −3 = k. ( ) ( )x
g
x
g
f






−




 1 = l. ( ) ( )x
f
ax
f 





 +
−




 13 = 
Solutions: 
a. ( ) ( )xgxf + = ( ) ( )1213 323 −+++ xxx = 1132 233 −+++ xxx = 23 33 xx + = ( )13 2 +xx 
 
b. ( ) ( )xgxf − = ( ) ( )1213 323 −−++ xxx = 1213 323 +−++ xxx = 1132 233 +++− xxx = 23 23 ++− xx 
 
c. ( )( ) 3
2
+xg
xf = ( )
( ) 312
132
3
23
+−
++
x
xx = ( )
312
132
3
23
+−
++
x
xx = ( )
22
132
3
23
+
++
x
xx = ( )
( )12
132
3
23
+/
++/
x
xx = 
1
13
3
23
+
++
x
xx 
 
d. ( ) ( )xgxf ⋅2 = ( ) ( )12132 323 −⋅++ xxx = ( )123622 32536 −+−+− xxxxx = ( )13622 2356 −−++ xxxx 
 
e. ( )( ) ( )( )xgxf 25 − = ( ) ( )122135 323 −−++ xxx = 445155 323 +−++ xxx = 915 23 ++ xx 
 
f. ( ) ( )xfxg +
1 = 13
12
1 23
3
+++
−
xx
x
 = 
1
13
12
1 23
3
++
+
−
xx
x
 = ( ) ( ) ( )
12
121311
3
323
−
−⋅+++⋅
x
xxx 
 
= 
12
123621
3
32536
−
−+−+−+
x
xxxxx = 
12
362
3
2356
−
−++
x
xxxx = ( )
12
362
3
342
−
−++
x
xxxx 
Mastering Algebra - Advanced Level 2.2 Math Operations Involving Functions of Real Variables 
Hamilton Education Guides 125 
 
g. ( )( )xgf 32 − = ( ) ( )123132 323 −−++ xxx = 36262 323 +−++ xxx = 564 23 ++− xx 
 
h. ( ) 42 −





x
g
f = ( ) 4
12
132
3
23
−
−
++
x
xx = 
1
4
12
262
3
23
−
−
++
x
xx = ( )[ ] ( )[ ]( ) 112
1241262
3
323
⋅−
−⋅−+⋅++
x
xxx 
 
= 
12
48262
3
323
−
+−++
x
xxx = 
12
42682
3
233
−
+++−
x
xxx = 
12
666
3
23
−
++−
x
xx = ( )
12
16
3
23
−
++−
x
xx 
 
i. ( ) ( )xgxf 5+ = ( ) ( )12513 323 −+++ xxx = 51013 323 −+++ xxx = 4311 23 −+ xx 
 
j. ( )( )xgf −3 = ( ) ( )12133 323 −−++ xxx = 12393 323 +−++ xxx = 13923 233 +++− xxx = 49 23 ++ xx 
 
k. ( ) ( )x
g
x
g
f






−




 1 = 
12
1
12
13
33
23
−
−
−
++
xx
xx = 
12
113
3
23
−
−++
x
xx = 
12
3
3
23
−
+
x
xx = ( )
12
3
3
2
−
+
x
xx 
 
l. ( ) ( )x
f
ax
f 





 +
−




 13 = 
13
1
13
3
2323 ++
+
−
++ xx
a
xx
 = ( )
13
13
23 ++
+−
xx
a = 
13
13
23 ++
−−
xx
a = 
13
2
23 ++
−
xx
a 
Example 2.2-2: Given ( )xf = xx 52 + and ( )xg = 12 −x , find 
a. ( ) ( )xgxf 3+ = b. ( ) ( )xgxf − = c. ( )( ) 1
3
+xg
xf = d. ( ) ( )xgxf ⋅4 = 
e. ( )( ) ( )( )xgxf 23 − = f. ( ) ( )xfxg +
1 = g. ( )( )xgf 52 − = h. ( )( )xgf −3 = 
Solutions: 
a. ( ) ( )xgxf + = ( ) ( )1252 −++ xxx = 1252 −++ xxx = 172 −+ xx 
 
b. ( ) ( )xgxf − = ( ) ( )1252 −−+ xxx = 1252 +−+ xxx = 132 ++ xx 
 
c. ( )( ) 1
3
+xg
xf = ( ) 112
52
+−
+
x
xx = 
112
52
+−
+
x
xx = ( )
x
xx
/
+/
2
5 = 
2
5+x 
 
d. ( ) ( )xgxf ⋅4 = ( ) ( )1254 2 −⋅+ xxx = ( )xxxx 51024 223 −+− = ( )xxx 5924 23 −+ = xxx 20368 23 −+ 
 
e. ( )( ) ( )( )xgxf 23 − = ( ) ( )12253 2 −−+ xxx = 24153 2 +−+ xxx = 2113 2 ++ xx 
 
f. ( ) ( )xfxg +
1 = ( )xx
x
5
12
1 2 ++
−
 = 
1
5
12
1 2 xx
x
+
+
−
 = ( ) ( )
12
1251 2
−
−⋅++
x
xxx = 
12
51021 223
−
−+−+
x
xxxx 
 
Mastering Algebra - Advanced Level 2.2 Math Operations Involving Functions of Real Variables 
Hamilton Education Guides 126 
 = ( )
12
151102 23
−
+−−+
x
xxx = 
12
1592 23
−
+−+
x
xxx 
 
g. ( )( )xgf 52 − = ( ) ( )12552 2 −−+ xxx = 510102 2 +−+ xxx = 52 2 +x 
 
h. ( )( )xgf −3 = ( ) ( )1253 2 −−+ xxx = 12153 2 +−+ xxx = 1133 2 ++ xx 
 
Example 2.2-3: Let ( ) 1+= xxf and ( ) xxxg += 2 . Find and simplify the following expressions. 
a. ( )( )1−+ gf = b. ( )( )2fg − = c. ( )2−





g
f = 
d. ( )( )2agf + = e. ( )10−





f
g = f. ( )( )3gf ⋅ = 
g. ( ) ( )22 gf − = h. ( )( )3−− fg = i. ( ) ( )22 fg − = 
j. ( )9
g
f = k. ( )( )9gf ⋅ = l. ( )a
f
g
− = 
Solutions: 
a. ( )( )xgf + = ( ) ( )xxx +++ 21 = 122 ++ xx therefore ( )( )1−+ gf = ( ) ( ) 1121 2 +−×+− = 121 +− = 0 
 
b. ( )( )xfg − = ( ) ( )12 +−+ xxx = 12 −−+ xxx = 12 −x therefore ( )( )2fg − = 122 − = 14− = 3 
 
c. ( )x
g
f





 = 
xx
x
+
+
2
1 = ( )1
1
+
+
xx
x = 
x
1 therefore ( )2−





g
f = 
2
1
− = 5.0− 
 
d. ( )( )xgf + = ( ) ( )xxx +++ 21 = 122 ++ xx therefore ( )( )2agf + = ( ) ( ) 12 222 +×+ aa = 12 24 ++ aa 
 
e. ( )x
f
g





 = 
1
2
+
+
x
xx = ( )
1
1
+
+
x
xx = 
1
x = x therefore ( )10−





f
g = 10− 
 
f. ( )( )xgf ⋅ = ( ) ( )xxx ++ 21 = xxxx +++ 223 = xxx ++ 23 2 thus ( )( )3gf ⋅ = 3323 23 +⋅+ = 48 
 
g. ( ) ( )xgxf − = ( ) ( )xxx +−+ 21 = xxx −−+ 21 = 12 +− x thus ( ) ( )22 gf − = ( ) 12 2 +− = 14+− = 3− 
 
h. ( )( )xfg − = ( ) ( )12 +−+ xxx = 12 −−+ xxx = 12 −x therefore ( )( )3−− fg = ( ) 13 2 −− = 19 − = 8 
 
i. ( ) ( )xfxg − = ( ) ( )12 +−+ xxx = 12 −−+ xxx = 12 −x therefore ( ) ( )22 fg − = 122 − = 14− = 3 
 
 
 
 
Mastering Algebra - Advanced Level 2.2 Math Operations Involving Functions of Real Variables 
Hamilton Education Guides 127 
j. ( )x
g
f = 
xx
x
+
+
2
1 = ( )1
1
+
+
xx
x = 
x
1 therefore ( )9
g
f = 
9
1 = 11.0 
 
k. ( )( )xgf ⋅ = ( ) ( )xxx ++ 21 = xxxx +++ 223 = xxx ++ 23 2 thus ( )( )9gf ⋅ = 9929 23 +⋅+ = 900 
 
l. ( )x
f
g = 
1
2
+
+
x
xx = ( )
1
1
+
+
x
xx = 
1
x = x therefore( )a
f
g
− = a− 
 
2.2.1 Determining Odd and Even Functions 
A function ( )xf is called an odd function if and only if: 
( )xf − = ( )xf− for all x 
A function ( )xf is called an even function if and only if: 
( )xf − = ( )xf for all x 
The following examples show how to identify odd or even functions: 
Example 2.2-4: State which of the following functions are odd or even. 
a. ( )xf = 2x b. ( )xf = 15 +x c. ( )xf = ( )1−xx 
d. ( )xf = 
x
x 1− e. ( )xf = 3x f. ( )xf = x+1 
g. ( )xf = 
x
x
−1
3
 h. ( )xf = 
x
x
+1
2
 i. ( )xf = 13 23 ++ xx 
Solutions: 
a. Given ( )xf = 2x compute ( )xf − and ( )xf− , i.e., ( )xf − = ( )2x− = 2x and ( )xf− = 2x− . Since 
( )xf − = ( )xf the function ( )xf = 2x is an even function. 
b. Given ( )xf = 15 +x compute ( )xf − and ( )xf− , i.e., ( )xf − = ( ) 15 +− x = 15 +− x and ( )xf− 
= ( )15 +− x = 15 −− x . Since ( )xf − ≠ ( )xf− and ( )xf − ≠ ( )xf the function ( )xf = 15 +x is 
neither an odd nor an even function. 
c. Given ( )xf = ( )1−xx = xx −2 compute ( )xf − and ( )xf− , i.e., ( )xf − = ( ) ( )xx −−− 2 = xx +2 and 
( )xf− = ( )xx −− 2 = xx +− 2 . Since ( )xf − ≠ ( )xf− and ( )xf − ≠ ( )xf the function ( )xf = 15 +x 
is neither an odd nor an even function. 
d. Given ( )xf = 
x
x 1− = 
x
x 12 − compute ( )xf − and ( )xf− , i.e., ( )xf − = ( )
x
x
−
−− 12 = 
x
x
−
−12 
= 
x
x 12 +− and ( )xf− = 





 −
−
x
x 12 = 
x
x 12 +− . Since ( )xf − = ( )xf− the function ( )xf = 
x
x 1− is 
an odd function. 
e. Given ( )xf = 3x compute ( )xf − and ( )xf− , i.e., ( )xf − = ( )3x− = 3x− and ( )xf− = 3x− . Since 
Mastering Algebra - Advanced Level 2.2 Math Operations Involving Functions of Real Variables 
Hamilton Education Guides 128 
( )xf − = ( )xf− the function ( )xf = 3x is an odd function. 
f. Given ( )xf = x+1 compute ( )xf − and ( )xf− , i.e., ( )xf − = x−+1 = x+1 and ( )xf− 
= ( )x+− 1 = x−−1 . Since ( )xf − = ( )xf the function ( )xf = x+1 is an even function. 
g. Given ( )xf = 
x
x
−1
3
 compute ( )xf − and ( )xf− , i.e., ( )xf − = ( )
x
x
−−
−
1
3
 = 
x
x
−
−
1
3
 and ( )xf− 
= 







−
−
x
x
1
3
 = 
x
x
−
−
1
3
. Since ( )xf − = ( )xf− the function ( )xf = 
x
x
−1
3
 is an odd function. 
h. Given ( )xf = 
x
x
+1
2
 compute ( )xf − and ( )xf− , i.e., ( )xf − = ( )
x
x
−+
−
1
2
 = 
x
x
+1
2
 and ( )xf− 
= 







+
−
x
x
1
2
 = 
x
x
+
−
1
2
. Since ( )xf − = ( )xf the function ( )xf = 
x
x
+1
2
 is an even function. 
i. Given ( )xf = 13 23 ++ xx compute ( )xf − and ( )xf− , i.e., ( )xf − = ( ) ( ) 13 23 +−+− xx = 13 23 ++− xx 
and ( )xf− = ( )13 23 ++− xx = 13 23 −−− xx . Since ( )xf − ≠ ( )xf− and ( )xf − ≠ ( )xf the function 
( )xf = 13 23 ++ xx is neither an odd nor an even function. 
In the next section we will learn how to find the composition of two or more functions shown as 
gf o = ( )( )xgf o = ( )( )xgf and hgf oo = ( )( )xhgf oo = ( )( )( )xhgf . 
Section 2.2 Practice Problems – Math Operations Involving Functions of Real Variables 
1. Given ( )xf = 532 +− xx and ( )xg = 22x , find 
 a. ( ) ( )xgxf +2 = b. ( ) ( )xgxf 3− = c. ( ) ( )xg
x
xf 5+ = 
 d. ( ) ( )xgxf ⋅3 = e. ( ) ( )xgxf 53 − = f. ( ) ( )xfxg 2
3
+ = 
 g. ( )( )xgf 23 − = h. ( ) 2
4
−





x
g
f = i. ( )( ) xxg
xf
+
− 2 = 
2. Let ( ) 22 += xxf and ( ) 52 += xxg . Find and simplify the following expressions. 
 a. ( )( )2−+ gf = b. ( )( )0fg − = c. ( )1−





g
f = 
 d. ( )( )02gf + = e. ( )2−





f
g = f. ( )( )2gf ⋅ = 
3. State which of the following functions are odd or even. 
 a. ( )xf = 1−x b. ( )xf = 16 +x c. ( )xf = ( )12 −xx 
 d. ( )xf = 
x
x 12 + e. ( )xf = 31 x+ f. ( )xf = 3+x 
 g. ( )xf = 
x
x
21
2
−
 h. ( )xf = 
x
x
+
−
1
1 i. ( )xf = 224 −+ xx 
Mastering Algebra - Advanced Level 2.3 Composite Functions of Real Variables 
Hamilton Education Guides 129 
2.3 Composite Functions of Real Variables 
It is sometimes useful to form a new function from two or more given functions. Composite 
functions are generally shown by a special symbol “ o” as ( )( )xgf o or gf o . Note that in math 
books the composition of two or three functions represented by ( )xf , ( )xg , and ( )xh is generally 
shown in one of the following forms: ( )xC = ( )( )xgf = ( )( )xgf o = gf o or ( )xC = ( )( )( )xhgf = 
( )( )xhgf oo = hgf oo . In this section, we will learn how to find composite functions using two 
or three functions. 
Case I Computing Composite Functions Using ( )xf and ( )xg 
Given ( ) 122 ++= xxxf and ( ) 3−= xxg , then ( )( )xgf is equal to 
( )( )xgf = ( )3−xf = ( )[ ] ( ) 122 ++ xgxg = ( ) ( ) 1323 2 +−+− xx = 162692 +−+−+ xxx = 442 +− xx 
The new function ( )( )xgf is called the composite function. Composite functions are obtained 
from different classes of ( )xf and ( )xg functions as shown in the following examples. 
Example 2.3-1: Find the composite function ( )( )xgf = ( )( )xgf o = gf o for the following ( )xf 
and ( )xg functions: 
a. ( ) 1+= xxf ; ( ) 2xxg = b. ( ) 3xxf = ; ( ) 1−= xxg c. ( )
1
1
2 −
=
x
xf ; ( ) 13 += xxg 
d. ( ) 62 += xxf ; ( ) 3xxg = e. ( ) xxxf += 2 ; ( ) 1+= xxg f. ( ) xxf = ; ( ) 12 += xxg 
g. ( ) 5xxf = ; ( )
1
1
+
=
x
xg h. ( )
1
11
−
+=
xx
xf ; ( ) 3xxg = i. ( ) 12 += xxf ; ( ) xxg = 
Solutions: 
a. Given ( ) 1+= xxf and ( ) 2xxg = , substitute ( )xg in place of x in ( )xf , i.e., 
( )( )xgf = ( ) 1+xg = 12 +x 
b. Given ( ) 3xxf = and ( ) 1−= xxg , substitute ( )xg in place of x in ( )xf , i.e., 
( )( )xgf = ( )3xg = ( ) 11 3 +−x 
c. Given ( )
1
1
2 −
=
x
xf and ( ) 13 += xxg , substitute ( )xg in place of x in ( )xf , i.e., 
( )( )xgf = 
( )[ ] 1
1
2 −xg
 = 
( ) 11
1
23 −+x
 = 
112
1
36 −++ xx
 = 36 2
1
xx +
 = ( )2
1
33 +xx
 
d. Given ( ) 62 += xxf and ( ) 3xxg = , substitute ( )xg in place of x in ( )xf , i.e., 
( )( )xgf = ( ) 62 +xg = 62 3 +x = ( )32 3 +x 
e. Given ( ) xxxf += 2 and ( ) 1+= xxg , substitute ( )xg in place of x in ( )xf , i.e., 
( )( )xgf = ( )[ ] ( )xgxg +2 = ( ) 11 2 +++ xx = 11 +++ xx 
f. Given ( ) xxf = and ( ) 12 += xxg , substitute ( )xg in place of x in ( )xf , i.e., 
Mastering Algebra - Advanced Level 2.3 Composite Functions of Real Variables 
Hamilton Education Guides 130 
( )( )xgf = ( )xg = 12 +x 
g. Given ( ) 5xxf = and ( )
1
1
+
=
x
xg , substitute ( )xg in place of x in ( )xf , i.e., 
( )( )xgf = ( )[ ]5xg = 
5
1
1






+x
 = 
( )51
1
+x
 
h. Given ( )
1
11
−
+=
xx
xf and ( ) 3xxg = , substitute ( )xg in place of x in ( )xf , i.e., 
( )( )xgf = ( ) ( ) 1
11
−
+
xgxg
 = 
1
11
33 −
+
xx
 = ( )
( )1
1
33
33
−
+−
xx
xx = 
( )1
1
33
33
−
+−
xx
xx = ( )1
12
33
3
−
−
xx
x 
i. Given ( ) 12 += xxf and ( ) xxg = , substitute ( )xg in place of x in ( )xf , i.e., 
( )( )xgf = ( )[ ] 12 +xg = ( ) 12 +x = 1+x 
Example 2.3-2: Find the composite function ( )( )xfg = ( )( )xfg o = fg o for the following ( )xg 
and ( )xf functions: 
a. ( ) 13 += xxg ; ( )
x
xf 1= b. ( ) 53 2 += xxg ; ( ) 15 += xxf c. ( ) 12 −= xxg ; ( ) 1+= xxf 
d. ( ) 13 −= xxg ; ( )
3
1
x
xf = e. ( )
2
1
1
1
xx
xg −
+
= ; ( ) xxf = f. ( ) x
x
xg 5
12
1
+
+
= ; ( ) 3xxf = 
Solutions: 
a. Given ( ) 13 += xxg and ( )
x
xf 1= , substitute ( )xf in place of x in ( )xg , i.e., 
( )( )xfg = ( )[ ] 13 +xf = 11
3
+





x
 = 113 +x
 
b. Given ( ) 53 2 += xxg and ( ) 15 += xxf , substitute( )xf in place of x in ( )xg , i.e., 
( )( )xfg = ( )[ ] 53 2 +xf = ( ) 5153 2 ++x = ( ) 5110253 2 +++ xx = 533075 2 +++ xx = 83075 2 ++ xx 
c. Given ( ) 12 −= xxg and ( ) 1+= xxf , substitute ( )xf in place of x in ( )xg , i.e., 
( )( )xfg = ( )[ ] 12 −xf = ( ) 11 2 −+x = 11−+x = x 
d. Given ( ) 13 −= xxg and ( )
3
1
x
xf = , substitute ( )xf in place of x in ( )xg , i.e., 
( )( )xfg = ( ) 13 −xf = 113 3 −⋅ x
 = 133 −x
 
e. Given ( )
2
1
1
1
xx
xg −
+
= and ( ) xxf = , substitute ( )xf in place of x in ( )xg , i.e., 
( )( )xfg = ( ) ( )[ ]2
1
1
1
xfxf
−
+
 = 
( )2
1
1
1
xx
−
+
 = 
xx
1
1
1
−
+
 = 
( )1
1
+
+−
xx
xx 
f. Given ( ) x
x
xg 5
12
1
+
+
= and ( ) 3xxf = , substitute ( )xf in place of x in ( )xg , i.e., 
Mastering Algebra - Advanced Level 2.3 Composite Functions of Real Variables 
Hamilton Education Guides 131 
( )( )xfg = ( ) ( )xfxf 512
1
+
+
 = 33 512
1 x
x
+
+
 
Note that changing the order in which functions are composed in most cases change the result. 
For example, given ( ) 12 += xxf and ( ) 3xxg = , then 
( )( ) ( )( )xgfxgf =o is equal to ( )( )xgf = ( )[ ] 12 +xg = ( ) 123 +x = 16 +x where as 
( )( ) ( )( )xfgxfg =o is equal to ( )( )xfg = ( )[ ]3xf = ( )32 1+x . 
However, given ( )
2
1
x
xf = and ( )
x
xg 1= result in having the same ( )( )xgf and ( )( )xfg , i.e., 
( )( ) ( )( )xgfxgf =o is equal to ( )( )xgf = 
( )[ ]2
1
xg
 = 
( )21
1
x
 = 
2
1
1
x
 = 
2
1
1
1
x
 = 
11
1 2
⋅
⋅ x = 
1
2x = 2x and 
( )( ) ( )( )xfgxfg =o is equal to ( )( )xfg = ( )xf
1 = 
2
1
1
x
 = 
2
1
1
1
x
 = 
11
1 2
⋅
⋅ x = 
1
2x = 2x . 
The following examples further illustrate this point. 
Example 2.3-3: Given the following ( )xf and ( )xg functions, find ( )( )xgf and ( )( )xfg . 
a. ( )
x
xxf 13 += and ( ) xxg = b. ( ) 52 ++= xxxf and ( ) 1+= xxg 
c. ( ) 12 −= xxf and ( )
3
1
x
xg = d. ( )
52
1
+
=
x
xf and ( ) 1−= xxg 
Solutions: 
a. Given ( )
x
xxf 13 += and ( ) xxg = substitute ( )xg in place of x in ( )xf to obtain ( )( )xgf . 
( )( )xgf = ( )[ ] ( )xgxg
13 + = ( )
x
x 1
3
+ = 
x
x 1
3
2
1
+




 = 
x
x 1
32
1
+
× = 
x
x 12
3
+ = 
x
x 13 + 
Next, substitute ( )xf in place of x in ( )xg to obtain ( )( )xfg . 
( )( )xfg = ( )xf = 
x
x 13 + 
d. Given ( ) 52 ++= xxxf and ( ) 1+= xxg substitute ( )xg in place of x in ( )xf to obtain ( )( )xgf . 
( )( )xgf = ( )[ ] ( ) 52 ++ xgxg = ( ) ( ) 511 2 ++++ xx = 51122 +++++ xxx = 732 ++ xx 
Next, substitute ( )xf in place of x in ( )xg to obtain ( )( )xfg . 
( )( )xfg = ( ) 1+xf = ( ) 152 +++ xx = 62 ++ xx 
c. Given ( ) 12 −= xxf and ( )
3
1
x
xg = substitute ( )xg in place of x in ( )xf to obtain ( )( )xgf . 
( )( )xgf = ( )[ ] 12 −xg = 11
2
3
−





x
 = 116 −x
 
Next, substitute ( )xf in place of x in ( )xg to obtain ( )( )xfg . 
Mastering Algebra - Advanced Level 2.3 Composite Functions of Real Variables 
Hamilton Education Guides 132 
( )( )xfg = 
( )[ ]3
1
xf
 = ( )32 1
1
−x
 
d. Given ( )
52
1
+
=
x
xf and ( ) 1−= xxg substitute ( )xg in place of x in ( )xf to obtain ( )( )xgf . 
( )( )xgf = ( ) 52
1
+xg
 = 
512
1
+−x
 
Next, substitute ( )xf in place of x in ( )xg to obtain ( )( )xfg . 
( )( )xfg = ( ) 1−xf = 1
52
1
−
+x
 = 
52
521
+
−−
x
x = 
52
42
+
−−
x
x 
Example 2.3-4: Given ( ) 353 −+= xxxf and ( ) 111
23
−+=
xx
xg , find 
a. ( )0f and ( )1−g b. ( )1−f and ( )3g c. ( )2af and ( )10g 
d. 





k
f 1 and ( )kg 2− e. ( )3−f and ( )2g f. ( )nf 2 and ( )0g 
Solutions: 
a. ( )0f = ( ) 30503 −⋅+ = 3− and ( )1−g = 
( ) ( )
1
1
1
1
1
23
−
−
+
−
 = 1
1
1
1
1
−+
−
 = 111 −+− = 1− 
 
b. ( )1−f = ( ) ( ) 3151 3 −−⋅+− = 351 −−− = 9− and ( )3g = 1
3
1
3
1
23
−+ = 1
9
1
27
1
−+ = 852.0− 
 
c. ( )2af = ( ) ( ) 35 232 −⋅+ aa = 35 26 −+ aa and ( )10g = 1
10
1
10
1
23
−+ = 1
100
1
1000
1
−+ = 989.0− 
 
d. 





k
f 1 = 3151
3
−




 ⋅+





kk
 = 3513 −+ kk
 and ( )kg 2− = 
( ) ( )
1
2
1
2
1
23
−
−
+
− kk
 = 1
4
1
8
1
23 −+− kk
 
 
e. ( )3−f = ( ) ( ) 3353 3 −−⋅+− = 31527 −−− = 45− and ( )2g = 1
2
1
2
1
23
−+ = 1
4
1
8
1
−+ = 625.0− 
 
f. ( )nf 2 = ( ) ( ) 3252 3 −⋅+ nn = 3108 3 −+ nn and ( )0g = 1
0
1
0
1
23
−+ = 1
0
1
0
1
−+ since division by 
zero is not defined the function ( )xg at 0=x approaches infinity. 
Example 2.3-5: Given ( ) 3+= xxf and ( ) 122 ++= xxxg , find 
a. ( )( )0gf b. ( )( )0fg c. ( )( )1−gf d. ( )( )agf 
e. ( )( )afg 10 f. ( )( )1+xfg g. ( )( )2xgf h. ( )( )2fg 
Solutions: 
First - Find ( )( )xgf and ( )( )xfg , i.e., 
( )( )xgf = ( ) 3+xg = ( ) 3122 +++ xx = 422 ++ xx 
Mastering Algebra - Advanced Level 2.3 Composite Functions of Real Variables 
Hamilton Education Guides 133 
 
( )( )xfg = ( )[ ] ( ) 122 +⋅+ xfxf = ( ) ( ) 1323 2 ++++ xx = 162962 +++++ xxx = 1682 ++ xx 
Second - Find ( )( )xgf or ( )( )xfg for the specific values given. 
a. ( )( )0gf = ( ) 40202 +⋅+ = 400 ++ = 4 
 
b. ( )( )0fg = ( ) 160802 +⋅+ = 1600 ++ = 16 
 
c. ( )( )1−gf = ( ) ( ) 4121 2 +−⋅+− = 421 +− = 3 
 
d. ( )( )agf = ( ) 422 +⋅+ aa = 422 ++ aa 
 
e. ( )( )afg 10 = ( ) ( ) 1610810 2 +⋅+ aa = 1680100 2 ++ aa 
 
f. ( )( )1+xfg = ( ) ( ) 16181 2 ++++ xx = 1688122 +++++ xxx = 25102 ++ xx 
 
g. ( )( )2xgf = ( ) ( ) 42 222 +⋅+ xx = 42 24 ++ xx 
 
h. ( )( )2fg = ( ) 162822 +⋅+ = 16164 ++ = 36 
Note that ( )( )xgf is defined only for values of x where ( )xg is defined. Similarly, ( )( )xfg is 
defined only for values of x where ( )xf is defined. For example, if ( )
2
1
x
xf = and ( )
x
xg 1= , then 
( )( )xgf = 
( )[ ]2
1
xg
 = 
( )21
1
x
 = 
2
1
1
x
 = 
2
1
1
1
x
 = 
11
1 2
⋅
⋅ x = 
1
2x = 2x and 
( )( )xfg = ( )xf
1 = 
2
1
1
x
 = 
2
1
1
1
x
 = 
11
1 2
⋅
⋅ x = 
1
2x = 2x . 
Therefore, we might expect that ( )( )xgf = 2x at 0=x be equal to zero, i.e., ( )( ) 000 2 ==gf . 
However, in fact, ( )( )0gf is undefined because ( )
x
xg 1= at 0=x is undefined. Similarly, we might 
expect ( )( )xfg at 0=x be equal to zero, i.e., ( )( ) 000 2 ==fg . Whereas, in fact, ( )( )0fg is again 
undefined because ( )
2
1
x
xf = at 0=x is undefined. 
 
 
 
 
 
 
 
Mastering Algebra - Advanced Level 2.3 Composite Functions of Real Variables 
Hamilton Education Guides 134 
Case II Computing Composite Functions Using Three Functions ( )xf , ( )xg , and ( )xh 
The above process of forming a composite function from two functions can be extended to three 
functions as shown in the following examples. 
Example 2.3-6: Given the following ( )xf , ( )xg , and ( )xh functions, find the composite function 
( )( )( )xhgf = ( )( )xhgf oo = hgf oo 
a. ( ) xxf 2= , ( ) 1+= xxg , and ( ) 2xxh = b. ( ) 1+= xxf , ( ) xxg 4= , and ( ) 3xxh = 
c. ( ) 23 += xxf , ( )
1
1
+
=
x
xg , and ( ) 5=xh d. ( ) 1+= xxf , ( ) 3xxg = , and ( ) 12 ++= xxxh 
e. ( ) ( )51+= xxf , ( )
1
1
−
=
x
xg , and ( ) xxh = f. ( ) 1+= xxf , ( ) 23 += xxg , and ( ) 1−= xxh 
Solutions: 
a. Given ( ) xxf 2= , ( ) 1+= xxg , and ( ) 2xxh = first substitute ( )xg in place of x in ( )xf to obtain 
gf o = ( )( )xgf . 
gf o = ( )( )xgf = ( )xg2 = ( )12 +x 
Next, substitute ( )xh in place of x in ( )( )xgf to obtain hgf oo = ( )( )( )xhgf . 
hgf oo = ( )( )( )xhgf = ( )( )12 +xh = ( )12 2 +x = 22 2 +x 
b. Given ( ) 1+= xxf , ( ) xxg 4= , and ( ) 3xxh = first substitute ( )xg in place of x in ( )xf to obtain 
gf o = ( )( )xgf . 
gf o = ( )( )xgf = ( ) 1+xg = 14 +x 
Next, substitute ( )xh in place of x in ( )( )xgf to obtain hgf oo = ( )( )( )xhgf . 
hgf oo = ( )( )( )xhgf = ( ) 14 +xh = 14 3 +x 
c. Given ( ) 23 += xxf ,( )
1
1
+
=
x
xg , and ( ) 5=xh first substitute ( )xg in place of x in ( )xf to obtain 
gf o = ( )( )xgf . 
gf o = ( )( )xgf = ( ) 23 +xg = 2
1
1 3
+





+x
 = 
( )
2
1
1
3
+
+x
 
Next, substitute ( )xh in place of x in ( )( )xgf to obtain hgf oo = ( )( )( )xhgf . 
hgf oo = ( )( )( )xhgf = 
( )[ ]
2
1
1
3
+
+xh
 = 
( )
2
15
1
3
+
+
 = 2
6
1
3
+ = 2
216
1
+ = 005.2 
d. Given ( ) 1+= xxf , ( ) 3xxg = , and ( ) 12 ++= xxxh first substitute ( )xg in place of x in ( )xf to 
obtain gf o = ( )( )xgf . 
gf o = ( )( )xgf = ( ) 1+xg = 13 +x 
Next, substitute ( )xh in place of x in ( )( )xgf to obtain hgf oo = ( )( )( )xhgf . 
hgf oo = ( )( )( )xhgf = ( )[ ] 13 +xh = ( ) 11 32 +++ xx 
Mastering Algebra - Advanced Level 2.3 Composite Functions of Real Variables 
 
Hamilton Education Guides 135 
e. Given ( ) ( )51+= xxf , ( )
1
1
−
=
x
xg , and ( ) xxh = first substitute ( )xg in place of x in ( )xf to 
obtain gf  = ( )( )xgf . 
gf  = ( )( )xgf = ( )[ ]51+xg = 
5
1
1
1





 +
−x
 
Next, substitute ( )xh in place of x in ( )( )xgf to obtain hgf  = ( )( )( )xhgf . 
hgf  = ( )( )( )xhgf = ( )
5
1
1
1






+
−xh
 = 
5
1
1
1






+
−x
 
f. Given ( ) 1+= xxf , ( ) 23 += xxg , and ( ) 1−= xxh , first substitute ( )xg in place of x in ( )xf to 
obtain gf  = ( )( )xgf . 
gf  = ( )( )xgf = ( ) 1+xg = ( ) 123 ++x = 33 +x 
Next, substitute ( )xh in place of x in ( )( )xgf to obtain hgf  = ( )( )( )xhgf . 
hgf  = ( )( )( )xhgf = ( )[ ] 33 +xh = ( ) 31 3 +−x 
Example 2.3-7: Given ( ) 5+= xxf , ( ) 132 ++= xxxg , and ( ) xxh = , find 
a. ( )( )( )xhgf b. ( )( )( )xhfg c. ( )( )( )xgfh 
d. ( )( )( )xfgh e. ( )( )( )xfhg f. ( )( )( )xghf 
Solutions: 
a. Find ( )( )xhg = ( )[ ] ( ) 132 ++ xhxh = 132 ++ xx then ( )( )( )xhgf = ( )( ) 5+xhg = ( ) 5132 +++ xx = 632 ++ xx 
 
b. Find ( )( )xhf = ( ) 5+xh = 5+x then ( )( )( )xhfg = ( )( )[ ] ( )( ) 132 ++ xhfxhf = ( ) ( ) 1535 2 ++++ xx 
 
= 115325102 +++++ xxx = 41132 ++ xx 
 
c. Find ( )( )xgf = ( ) 5+xg = ( ) 5132 +++ xx = 632 ++ xx then ( )( )( )xgfh = ( )( )xgf = 632 ++ xx 
 
d. Find ( )( )xfg = ( )[ ] ( ) 132 ++ xfxf = ( ) ( ) 1535 2 ++++ xx = 115325102 +++++ xxx = 41132 ++ xx 
 
then ( )( )( )xfgh = ( )( )xfg = 41132 ++ xx 
 
e. Find ( )( )xfh = ( )xf = 5+x then ( )( )( )xfhg = ( )( )[ ] ( )( ) 132 ++ xfhxfh = ( ) ( ) 1535 2 ++++ xx 
 
= 115325102 +++++ xxx = 41132 ++ xx 
 
f. Find ( )( )xgh = ( )xg = 132 ++ xx then ( )( )( )xghf = ( )( ) 5+xgh = ( ) 5132 +++ xx = 632 ++ xx 
 
Mastering Algebra - Advanced Level 2.3 Composite Functions of Real Variables 
 
Hamilton Education Guides 136 
Example 2.3-8: Given the functions ( )xf , ( )xg , and ( )xh in example 2.3-7, find 
a. ( )( )( )0hgf b. ( )( )( )1hgf c. ( )( )( )1−hfg 
d. ( )( )( )0hfg e. ( )( )( )afgh − f. ( )( )( )0fgh 
g. ( )( )( )2−fhg h. ( )( )( )2fhg i. ( )( )( )1ghf 
Solutions: 
a. Given ( )( )( )xhgf = 632 ++ xx then ( )( )( )0hgf = ( ) 60302 +×+ = 6 
 
b. Given ( )( )( )xhgf = 632 ++ xx then ( )( )( )1hgf = ( ) 61312 +×+ = 631 ++ = 10 
 
c. Given ( )( )( )xhfg = 41132 ++ xx then ( )( )( )1−hfg = ( ) ( ) 411131 2 +−×+− = 41131 +− = 29 
 
d. Given ( )( )( )xhfg = 41132 ++ xx then ( )( )( )0hfg = ( ) 4101302 +×+ = 41 
 
e. Given ( )( )( )xfgh = 41132 ++ xx then ( )( )( )afgh − = ( ) ( ) 41132 +−×+− aa = 41132 +− aa 
 
f. Given ( )( )( )xfgh = 41132 ++ xx then ( )( )( )0fgh = ( ) 4101302 +×+ = 41 
 
g. Given ( )( )( )xfhg = 41132 ++ xx then ( )( )( )2−fhg = ( ) ( ) 412132 2 +−×+− = 41264 +− = 19 
 
h. Given ( )( )( )xfhg = 41132 ++ xx then ( )( )( )2fhg = ( ) 4121322 +×+ = 41264 ++ = 71 
 
i. Given ( )( )( )xghf = 632 ++ xx then ( )( )( )1ghf = ( ) 61312 +×+ = 631 ++ = 10 
In order to solve some of the problems presented in calculus students are asked to separate 
composite functions into two other functions, i.e., ( )xf and ( )xg functions. The quickest way to 
solve these types of problems is by: 
First - identifying the basic math operations (such as addition, subtraction, division, power, and 
radical) that are used to form the composite function and 
Second - Choosing the correct ( )xf and ( )xg functions that satisfy the identified operations. 
For example, the function ( ) ( )( ) ( )32 1+== xxgfxC consists of two operations, i.e., the addition and 
the power operations. Therefore, let ( )xf be equal to ( ) 3xxf = and ( )xg be equal to 
( ) ( )12 += xxg , then ( ) ( )( ) ( )[ ] ( )323 1+=== xxgxgfxC . The following examples further illustrate 
how composite functions can be separated into two functions. 
Example 2.3-9: Given the composite functions below, find the ( )xf and ( )xg functions such 
that ( ) ( )( ) ( )xCxgfxgf == . 
a. ( )xC = 21 x+ b. ( )xC = 101 +
x
 c. ( )xC = 13 +x 
Mastering Algebra - Advanced Level 2.3 Composite Functions of Real Variables 
 
Hamilton Education Guides 137 
d. ( )xC = 
( )51
1
+x
 e. ( )xC = 
ax
a
+
 f. ( )xC = xx + 
g. ( )xC = 52 +x h. ( )xC = 
( )31
1
−x
 i. ( )xC = 
12
1
−+x
 
Solutions: 
a. The composite function ( ) 21 xxC += consist of addition and power operations. Therefore, let 
( ) xxf +=1 and ( ) 2xxg = then ( )( )xgf = ( )xg+1 = 21 x+ 
b. The composite function ( ) 101 +=
x
xC consist of addition and division operations. Therefore, let 
( ) 10+= xxf and ( )
x
xg 1= then ( )( )xgf = ( ) 10+xg = 101 +
x
 
c. The composite function ( ) 13 += xxC consist of addition, power, and radical operations. 
Therefore, let ( ) xxf = and ( ) 13 += xxg then ( )( )xgf = ( )xg = 13 +x 
d. The composite function ( )
( )51
1
+
=
x
xC consist of addition, division, and power operations. 
Therefore, let ( ) 5
1
x
xf = and ( ) 1+= xxg then ( )( )xgf = 
( )[ ]5
1
xg
 = 
( )51
1
+x
 
e. The composite function ( )
ax
axC
+
= consist of addition and division operations. Therefore, 
let ( )
x
axf = and ( ) axxg += then ( )( )xgf = ( )xg
a = 
ax
a
+
 
f. The composite function ( ) xxxC += consist of addition and radical operations. Therefore, let 
( ) xxxf += 2 and ( ) xxg = then ( )( )xgf = ( )[ ] ( )xgxg +2 = ( ) xx +2 = xx + 
g. The composite function ( ) 52 += xxC consist of addition and power operations. Therefore, let 
( ) 5+= xxf and ( ) 2xxg = then ( )( )xgf = ( ) 5+xg = 52 +x 
h. The composite function ( ) ( )31−= xxC consist of subtraction and power operations. Therefore, 
let ( ) 3xxf = and ( ) 1−= xxg then ( )( )xgf = ( )[ ]3xg = ( )31−x 
i. The composite function ( )
12
1
−+
=
x
xC consist of addition, subtraction, and power operations. 
Therefore, let ( )
1
1
−
=
x
xf and ( ) 2+= xxg then ( )( )xgf = 
( ) 1
1
−xg
 = 
12
1
−+x
 
Note that composite functions, in some cases, can be computed by writing different forms of 
Mastering Algebra - Advanced Level 2.3 Composite Functions of Real Variables 
 
Hamilton Education Guides 138 
( )xf or ( )xg functions. For example, the above composite function ( )( )xgf = 
12
1
−+x
 can also 
be obtained by selecting the following ( )xf and ( )xg functions: 
1. Let, ( )
1
1
−
=
x
xf and ( ) 2+= xxg , then ( )( )xgf = ( ) 1
1
−xg
 = 
12
1
−+x
, or 
2. Let, ( )
x
xf 1= and ( ) 12 −+= xxg , then ( )( )xgf = ( )xg
1 = 
12
1
−+x
 
Having learned about the composite functions, in the next sections we will learn how to identify 
one-to-one functions and compute the inverse of a function. 
Section 2.3 Practice Problems – Composite Functions of Real Variables 
1. Find the composite function ( )( )xgf = ( )( )xgf  for the following ( )xf and ( )xg functions. 
 a. ( ) 12 −=xxf ; ( ) 2xxg −= b. ( ) 52 += xxf ; ( ) 10+= xxg c. ( )
1
1
+
=
x
xf ; ( ) 3xxg = 
 d. ( ) 3−= xxf ; ( ) 2xxg −= e. ( ) 12 += xxf ; ( ) xxg = f. ( ) xxxf 10+= ; ( ) 3−= xxg 
2. Find the composite function ( )( )xfg = ( )( )xfg  for the following ( )xg and ( )xf functions. 
 a. ( ) 2xxg = ; ( )
x
xf 1−= b. ( ) 12 −= xxg ; ( ) xxf 3= c. ( ) 2+= xxg ; ( ) 5+= xxf 
 d. ( ) 12 += xxg ; ( )
2
1
x
xf = e. ( )
1
1
+
=
x
xg ; ( ) xxf −= f. ( ) x
x
xg +
−
=
32
1 ; ( ) xxf 3= 
3. Given ( ) 52 +−= xxf and ( ) 123 −+= xxxg , find 
 a. ( )0f and ( )1g b. ( )1f and ( )2g c. ( )kf and ( )kg − 
 d. ( )3f and ( )3−g e. ( )nf − and ( )ng 2 f. ( )1+nf and ( )0g 
4. Given ( ) 2−= xxf and ( ) 12 −= xxg , find 
 a. ( )( )1−gf b. ( )( )0fg c. ( )( )3−gf 
 d. ( )( )2fg e. ( )( )1−ngf f. ( )( )nfg 2 
5. Given the functions ( )( )( ) 12 23 +−= xxxhgf , ( )( )( ) 1−= xxhfg , and ( )( )( ) 22 +−= xxfgh , find 
 a. ( )( )( )2hgf b. ( )( )( )2−hgf c. ( )( )( )1−afgh 
 d. ( )( )( )1−hfg e. ( )( )( )12 +nhfg f. ( )( )( )0fgh 
Mastering Algebra - Advanced Level 2.4 One-to-One and Inverse Functions of Real Variables 
Hamilton Education Guides 139 
2.4 One-to-One and Inverse Functions of Real Variables 
In this section we will learn how to identify one-to-one functions. We will then discuss two 
different methods for finding the inverse of functions. 
2.4.1 One-to-One Functions 
A function is a one-to-one function if and only if for each value of x there is only one 
corresponding value of y . In other words, a function having two ordered pairs with different 
first coordinates and the same second coordinates is not a one-to-one function. To find out if a 
function is a one-to-one function we consider two cases. In the first case, the function is 
represented by a set of ordered pairs. In the second case, the function is represented in an 
equation form. Let’s find out how a one-to-one function for each of the stated cases is found. 
Case I The Function is Represented by a Set of Ordered Pairs 
To state if a function is a one-to-one function, for cases where the function is given by a set of 
ordered pairs, we simply check to see if for each value of x there is only one corresponding value 
of y . For example, the function ( )yxf , = ( ) ( ) ( ) ( )[ ]8,6,2,5,3,4,1,2 is a one-to-one function. 
However, the function ( )yxf , = ( ) ( ) ( )[ ]5,1,7,3,5,1− is not a one-to-one function because of the 
ordered pairs ( )5,1− and ( )5,1 , i.e., for 1±=x there is one value of 5=y . 
Example 2.4-1: State which of the following ordered paired functions are one-to-one: 
a. ( ) ( ) ( )[ ]9,7,6,2,4,1=f b. ( ) ( ) ( ) ( )[ ]9,6,7,3,6,2,4,2=f 
c. ( ) ( ) ( ) ( )[ ]6,3,3,1,0,2,1,3 −−−=f d. ( ) ( ) ( ) ( )[ ]15,9,10,8,7,4,5,2=f 
Solutions: 
a. The ordered pair function ( ) ( ) ( )[ ]9,7,6,2,4,1=f is a one-to-one function. Because each x value 
correspond to only one y value. 
b. The ordered pair function ( ) ( ) ( ) ( )[ ]9,6,7,3,6,2,4,2=f is not a one-to-one function. Because 
each x value does not correspond with a different y value, i.e., when 2=x the variable y is 
equal to 4 and 6 . 
c. The ordered pair function ( ) ( ) ( ) ( )[ ]6,3,3,1,0,2,1,3 −−−=f is a one-to-one function. Because 
each x value correspond to only one y value. 
d. The ordered pair function ( ) ( ) ( ) ( )[ ]15,9,10,8,7,4,5,2=f is a one-to-one function. Because each 
x value correspond to only one y value. 
Case II The Function is Represented by an Equation 
To state if a function is a one-to-one function, for cases where the function is given in an 
equation form, we can use one of the following two methods. 
First Method: 
Substitute different x values into the given equation and solve for the y value. If each x value 
result in having only one y value, then the function is one-to-one. For example, the equations 
Mastering Algebra - Advanced Level 2.4 One-to-One and Inverse Functions of Real Variables 
Hamilton Education Guides 140 
( ) 85 −= xxf , ( ) 1+= xxf , ( ) 2+= xxf , ( ) 52 5
1
+= xxf , ( )
3
2
−
+
=
x
xxf , ( ) 3xxf = , and ( ) ( ) 47 31 +−= xxf 
are one-to-one functions. This is because for every value of x there is only one value of y . On 
the other hand, functions such as ( ) 12 += xxf , ( ) 4xxf = , ( )
x
xf
+
=
1
1 , ( ) 21 xxf −= , ( ) ( )23−= xxf , 
( ) ( )( )63 +−= xxxf , ( ) 2xxf = , ( ) ( )5221 xxf −= , and ( ) 652 ++= xxxf are not one-to-one functions. 
This is because, as an example, at 1±=x or at 2±=x the function ( ) 12 += xxf is equal to 
( ) ( ) 1111 2 +=+±=xf 2= and ( ) ( ) 54112 2 =+=+±=xf . Thus, the ordered pairs are equal to 
( ) ( ) ( ) ( ){ }5,2,5,2,2,1,2,1 −− which indicate that the function ( ) 12 += xxf is not a one-to-one 
function. The following examples further illustrate this point. 
Example 2.4-2: State which of the following functions are one-to-one. 
a. ( ) 13 += xxf b. ( ) 2xxf = c. ( ) 210 xxf += 
d. ( ) 3xxf = e. ( ) 1+= xxf f. ( )
1
1
2 +
=
x
xf 
g. ( ) 52 +−= xxf h. ( ) 52 3 −= xxf i. ( ) xexf +=1 
Solutions: 
a. Given the function ( ) 13 +== xyxf , let’s find the y value by substituting few x values into the 
equation 13 += xy , i.e., 
at 0=x y = ( ) 103 +⋅ = 10+ = 1 at 1=x y = ( ) 113 +⋅ = 13+ = 4 
at 1−=x y = ( ) 113 +−⋅ = 13+− = 2− at 3=x y = ( ) 133 +⋅ = 19+ = 10 
at 5=x y = ( ) 153 +⋅ = 115+ = 16 at 7=x y = ( ) 173 +⋅ = 121+ = 22 
Therefore, the ordered pairs are equal to ( ) ( ) ( ) ( ) ( ) ( ){ }22,7,16,5,10,3,2,1,4,1,1,0 −− . Since for 
every x value there is only one corresponding y value the function ( ) 13 +== xyxf is a one-
to-one function. 
b. Given the function ( ) 2xyxf == , let’s find the y value by substituting few x values into the 
equation 2xy = , i.e., 
at 0=x y = 20 = 0 at 1=x y = 21 = 1 
at 1−=x y = ( )21− = 1 at 4=x y = 24 = 16 
at 4−=x y = ( )24− = 16 at 7=x y = 27 = 49 
Therefore, the ordered pairs are equal to ( ) ( ) ( ) ( ) ( ) ( ){ }49,7,16,4,16,4,1,1,1,1,0,0 −− . Since each x 
value does not correspond to only one y value the function ( ) 2xyxf == is not a one-to-one 
function. 
c. Given the function ( ) 210 xyxf +== , let’s find the y value by substituting few x values into the 
equation 210 xy += , i.e., 
at 0=x y = 2010+ = 010+ = 10 at 1=x y = 2110+ = 110+ = 11 
at 1−=x y = ( )2110 −+ = 110+ = 11 at 5=x y = 2510+ = 2510+ = 35 
at 5−=x y = ( )2510 −+ = 2510+ = 35 at 8=x y = 2810+ = 6410+ = 74 
Mastering Algebra - Advanced Level 2.4 One-to-One and Inverse Functions of Real Variables 
Hamilton Education Guides 141 
Therefore, the ordered pairs are equal to ( ) ( ) ( ) ( ) ( ) ( ){ }74,8,35,5,35,5,11,1,11,1,10,0 −− . Since 
each x value does not correspond to only one y value the function ( ) 210 xyxf +== is not a 
one-to-one function. 
d. Given the function ( ) 3xyxf == , let’s find the y value by substituting few x values into the 
equation 3xy = , i.e., 
at 0=x y = 30 = 0 at 1=x y = 31 = 1 
at 1−=x y = ( )31− = 1− at 3=x y = 33 = 27 
at 3−=x y = ( )33− = 27− at 4=x y = 34 = 64 
Therefore, the ordered pairs are equal to ( ) ( ) ( ) ( ) ( ) ( ){ }64,4,27,3,27,3,1,1,1,1,0,0 −−−− . Since for 
every x value there is only one corresponding y value the function ( ) 3xyxf == is a one-to-
one function. 
e. Given the function ( ) 1+== xyxf , let’s find the y value by substituting few x values into the 
given equation, i.e., 
at 0=x y = 10+ = 1 = 1 at 1=x y = 11+ = 2 = 2 
at 1−=x y = 11+−= 0 = 0 at 2=x y = 12+ = 3 = 3 
at 2−=x y = 12+− = 1− = 1 = 1 at 3−=x y = 13+− = 2− = 2 = 2 
Therefore, the ordered pairs are equal to ( ) ( ) ( ) ( ) ( ) ( ){ }2,3,1,2,3,2,0,1,2,1,1,0 −−− . Since each x 
value does not correspond to only one y value the function ( ) 1+== xyxf is not a one-to-
one function. 
f. Given the function ( )
1
1
2 +
==
x
yxf , let’s find the y value by substituting few x values into the 
given equation, i.e., 
at 0=x y = 
10
1
2 +
 = 
10
1
+
 = 
1
1 = 1 at 1=x y = 
11
1
2 +
 = 
11
1
+
 = 
2
1 = 5.0 
at 1−=x y = 
( ) 11
1
2 +−
 = 
11
1
+
 = 
2
1 = 5.0 at 5=x y = 
15
1
2 +
 = 
125
1
+
 = 
26
1 = 04.0 
at 5−=x y = 
( ) 15
1
2 +−
 = 
125
1
+
 = 
26
1 = 04.0 at 10=x y = 
110
1
2 +
 = 
1100
1
+
 = 
101
1 = 009.0 
Therefore, the ordered pairs are equal to ( ) ( ) ( ) ( ) ( ) ( ){ }009.0,10,04.0,5,04.0,5,5.0,1,5.0,1,1,0 −− . 
Since each x value does not correspond to only one y value the function ( )
1
1
2 +
==
x
yxf is 
not a one-to-one function. 
g. Given the function ( ) 52 +−= xxf , let’s find the y value by substituting few x values into the 
equation 52 +−= xy , i.e., 
at 0=x y = 520 +− = 52 +− = 52 + = 7 at 1=x y = 521 +− = 51 +− = 51 + = 6 
at 3=x y = 523 +− = 51 + = 6 at 5=x y = 525 +− = 53 + = 8 
Therefore, the ordered pairs are equal to ( ) ( ) ( ) ( ){ }8,5,6,3,6,1,7,0 . Since each x value does not 
corresponde to only one y value the function ( ) 52 +−== xyxf is not a one-to-one function. 
Mastering Algebra - Advanced Level 2.4 One-to-One and Inverse Functions of Real Variables 
Hamilton Education Guides 142 
h. Given the function ( ) 52 3 −== xyxf , let’s find the y value by substituting few x values into 
the equation 52 3 −= xy , i.e., 
at 0=x y = 502 3 −⋅ = 50− = 5− at 1=x y = 512 3 −⋅ = 52− = 3− 
at 1−=x y = ( ) 512 3 −−⋅ = 52−− = 7− at 3=x y = 532 3 −⋅ = 554− = 49 
at 3−=x y = ( ) 532 3 −−⋅ = 554−− = 59− at 4=x y = 542 3 −⋅ = 5128− = 123 
Therefore, the ordered pairs are equal to ( ) ( ) ( ) ( ) ( ) ( ){ }123,4,59,3,49,3,7,1,3,1,5,0 −−−−−− . Since 
for every x value there is only one corresponding y value the function ( ) 52 3 −== xyxf is a 
one-to-one function. 
i. Given the function ( ) xeyxf +== 1 , let’s find the y value by substituting few x values into the 
equation 13 += xy , i.e., 
at 0=x y = 01 e+ = 11+ = 2 at 1=x y = 11 e+ = 718.21+ = 718.3 
at 1−=x y = 11 −+ e = 
e
11+ = 
718.2
11+ = 37.1 at 2=x y = 21 e+ = 389.71+ = 389.8 
at 2−=x y = 21 −+ e = 
2
11
e
+ = 
389.7
11+ = 14.1 at 3=x y = 31 e+ = 08.201+ = 08.21 
Therefore, the ordered pairs are equal to ( ) ( ) ( ) ( ) ( ) ( ){ }08.21,3,14.1,2,389.8,2,37.1,1,718.3,1,2,0 −− . 
Since for every x value there is only one corresponding y value the function ( ) xeyxf +== 1 is 
a one-to-one function. 
Second Method: 
The second way of determining whether a function is one-to-one is by first graphing the function 
and then drawing a horizontal line through the graph. If the graph of the functions crosses the 
horizontal line only once, then the function is a one-to-one function. However, if the graph of 
the function crosses the horizontal line two or more times, then the function is not a one-to-one 
function. For example, 122 =+ yx ; 21 xy −= ; 2
4
1 xy = ; ( )23−= xy ; 31 +−= xy ; xy = are not 
one-to-one functions because a horizontal line crosses the graph of each function more than once. 
In summary, a function is a one-to-one function if and only if: 
a. For each value of x there is only one corresponding value of y , i.e., no two values of x result 
in the same value of y , or 
b. The graph of the function intersects a horizontal line only once. 
2.4.2 Inverse Functions 
To find the inverse of a function we consider two cases. In the first case, the function is 
represented by a set of ordered pairs. In the second case, the function is represented in an 
equation form. Let’s find out how the inverse of a function for each of the stated cases are found. 
Case I The Function is Represented by a Set of Ordered Pairs 
To find the inverse of a function, for cases where the function is given by a set of ordered pairs, 
we simply interchange the coordinates in the ordered pairs. For example, the inverse of the 
function ( ) ( ) ( )[ ]7,3,6,2,5,1=f is obtained by interchanging the x and y coordinates of each 
Mastering Algebra - Advanced Level 2.4 One-to-One and Inverse Functions of Real Variables 
Hamilton Education Guides 143 
ordered pair of f , i.e., ( ) ( ) ( )[ ]3,7,2,6,1,51 =−f . Observe that interchanging the coordinates in the 
ordered pairs result in having the domain of 1−f (read as “ f inverse of x ”) to be the same as the 
range of f , and the range of 1−f to be the same as the domain of f . Also note that invertability 
of a function is dependent upon the ordered pairs of the function. The ordered pairs of a function 
is invertible if and only if the function is a one-to-one function. In section 2.1 we stated that a 
function is defined as a relation in which each element of the domain is paired with only one 
element of the range. Therefore, if a function ( )xf has ordered pairs with different first 
coordinates and the same second coordinate, then that function is not a one-to-one function and is 
not invertible. 
Example 2.4-3: Find the inverse of the following ordered paired functions: 
a. ( ) ( ) ( )[ ]6,8,5,2,3,1=f b. ( ) ( ) ( ) ( )[ ]10,6,8,3,8,1,5,1=f 
c. ( ) ( ) ( ) ( )[ ]5,2,3,1,0,1,1,2 −−=f d. ( ) ( ) ( ) ( )[ ]25,10,16,8,12,4,10,3=f 
Solutions: 
a. By interchanging the x and y values we obtain 1−f = ( ) ( ) ( )[ ]8,6,2,5,1,3 
b. The ordered pair function ( ) ( ) ( ) ( )[ ]10,6,8,3,8,1,5,1=f does not have an inverse. Because each x 
value does not correspond with a different y value, i.e., when 1=x the variable y is equal to 
5 and 8 . 
c. By interchanging the x and y values we obtain 1−f = ( ) ( ) ( ) ( )[ ]2,5,1,3,1,0,2,1 −− 
d. By interchanging the x and y values we obtain 1−f = ( ) ( ) ( ) ( )[ ]10,25,8,16,4,12,3,10 
Case II The Function is Represented by an Equation 
To find the inverse of a function, for cases where the function is given in an equation form, we 
can use one of the following two methods. 
First Method: 
In the first method for a given function ( )f x the inverse of ( )f x is found by simply interchanging 
the x and y variables as shown in the following steps: 
Step 1: Check to see if the function is a one-to-one function. 
Step 2: Replace ( )xf by y and interchange the x variable with y and the y variable with x . 
Step 3: Solve the equation for y . Replace y by ( )xf 1− . 
Second Method: 
In the second method since the inverse of ( )f x , denoted by ( )xf 1− , is the unique function that 
satisfies the equation 
( )[ ]xff 1− = x for all x (1) 
we can use this equation to find the inverse of the function. The following show the steps for 
finding the inverse of a function such as ( )xf = 12 +x , where 0≥x : 
Step 1: Check to see if the function ( )xf = 12 +x is a one-to-one function.
Mastering Algebra - Advanced Level 2.4 One-to-One and Inverse Functions of Real Variables 
Hamilton Education Guides 144 
Step 2: Since the inverse of any function satisfies ( )[ ]xff 1− = x , use this equation - labeled as 
equation (1) - and replace x with ( )xf 1− wherever x is present in the function ( )xf , i.e., 
since ( )xf = 12 +x , then ( )[ ]xff 1− = ( )[ ] 121 +− xf . Label this equation as equation No. (2). 
Step 3: Equate the right hand side of the equations (1) and (2) with one another and solve for 
( )xf 1− , i.e., since ( )[ ]xff 1− = x and ( )[ ]xff 1− = ( )[ ] 121 +− xf ,then ( )[ ] 121 += − xfx 
; ( )[ ]211 xfx −=− ; ( )[ ]211 xfx −=− ; ( )xfx 11 −=− or ( ) 11 −=− xxf . 
Note: The symbol 1−f , used for denoting the inverse of a function, does not represent a negative 
exponent and should not be confused as an exponent. For example, the functions 
x
x
e
e
2
2 1=− , 
3
3 1
x
x =− , or ( )
2
12 1
+
=+ −
x
x where as 
f
f 11 ≠− . 
The following examples show the steps for finding the inverse of a function. 
Example 2.4-4: Use two methods to find the inverse of the following functions: 
a. ( ) 12 += xxf b. ( ) 22 −= xxf c. ( ) 8−−= xxf 
d. ( ) 31 xxf += e. ( ) 12 −= xxf , 2
1≥x f. ( ) 55 −= xxf 
g. ( ) 53 5 −= xxf h. ( ) 5
3
1
−= xxf i. ( ) 100+= xxf 
j. ( ) xxf 4.0= k. ( ) 31 +=
x
xf , 0≠x l. ( ) xxf −= 
m. ( ) 375 +−= xxf , 7≥x n. ( )
1
52
−
−
=
x
xxf , 1≠x o. ( ) xxf 37 −= 
Solutions: 
a. The function ( ) 12 += xxf is a one-to-one function because for each value of x there is only 
one corresponding value of y . For example, at 0=x ; 1=y or at 2=x ; 5=y , etc. 
First Method: Interchange the x variable with y and the y variable with x in the equation 
( ) 12 +== xyxf and solve for y . Next, replace y with ( )xf 1− to obtain: 
12 += yx ; yx 21=− ; 
2
2
2
1 yx
=
− ; yx =−
2
1 ; 
2
1−
=
xy ; ( )
2
11 −=− xxf 
Second Method: Since ( )[ ]xff 1− = x replace x with ( )xf 1− wherever x is present in the 
function ( ) 12 += xxf , i.e., ( )[ ] ( ) 12 11 +⋅= −− xfxff . 
Next, solve for ( )xf 1− by equating the right hand side of the equations ( )[ ] xxff =−1 and 
( )[ ] ( ) 12 11 +⋅= −− xfxff , i.e., ( ) 12 1 +⋅= − xfx ; ( )xfx 121 −=− ; ( )
2
2
2
1 1 xfx −
=
− ; ( )
2
11 −=− xxf 
 
Mastering Algebra - Advanced Level 2.4 One-to-One and Inverse Functions of Real Variables 
Hamilton Education Guides 145 
b. The function ( ) 22 −= xxf is not a one-to-one function. This is because each x value does not 
correspond to only one y value, i.e., at 1±=x ; 1−=y or at 3±=x ; 7=y , etc. Therefore, the 
function ( ) 22 −= xxf is not invertible. 
c. The function ( ) 8−−= xxf is a one-to-one function because for each value of x there is only 
one corresponding value of y . For example, at 1=x ; 9−=y or at 2=x ; 10−=y , etc. 
First Method: Interchange the x variable with y and the y variable with x in the equation 
( ) 8−−== xyxf and solve for y . Next, replace y with ( )xf 1− to obtain: 
8−−= yx ; yx −=+8 ; 8+=− xy ; 8−−= xy ; ( ) 81 −−=− xxf 
Second Method: Since ( )[ ]xff 1− = x replace x with ( )xf 1− wherever x is present in the 
function ( ) 8−−= xxf , i.e., ( )[ ] ( ) 811 −−= −− xfxff . 
Next, solve for ( )xf 1− by equating the right hand side of the equations ( )[ ] xxff =−1 and 
( )[ ] ( ) 811 −−= −− xfxff , i.e., ( ) 81 −−= − xfx ; ( )xfx 18 −−=+ ; ( ) 81 −−=− xxf 
d. The function ( ) 31 xxf += is a one-to-one function because for each value of x there is only 
one corresponding value of y . For example, at 1=x ; 2=y or at 1−=x ; 0=y , etc. 
First Method: Interchange the x variable with y and the y variable with x in the equation 
( ) 31 xyxf +== and solve for y . Next, replace y with ( )xf 1− to obtain: 
31 yx += ; 31 yx =− ; ( ) ( )3131 31 yx =− ; ( ) yx =− 311 ; 3 1−= xy ; ( ) 31 1−=− xxf 
Second Method: Since ( )[ ]xff 1− = x replace x with ( )xf 1− wherever x is present in the 
function ( ) 31 xxf += , i.e., ( )[ ] ( )[ ]311 1 xfxff −− += . 
Next, solve for ( )xf 1− by equating the right hand side of the equations ( )[ ] xxff =−1 and 
( )[ ] ( )[ ]311 1 xfxff −− += , i.e., ( )[ ]311 xfx −+= ; ( )[ ]311 xfx −=− ; ( ) ( )[ ] 3131 311 ×−=− xfx ; ( ) ( )xfx 1311 −=− 
 
( ) ( )3111 −=− xxf ; ( ) ( )31 1−=− xxf 
e. The function ( ) 12 −= xxf is a one-to-one function because for each value of 2
1≥x there is only 
 one corresponding value of y , i.e., at 2
1=x ; 0=y or at 2=x ; 3=y , etc. 
First Method: Interchange the x variable with y and the y variable with x in the equation 
( ) 12 −== xyxf and solve for y . Next, replace y with ( )xf 1− to obtain: 
12 −= yx ; ( )2112 −= yx ; ( ) 22 2112 ×−= yx ; 122 −= yx ; yx 212 =+ ; 
2
2
2
12 yx
=
+ ; yx =+
2
12 
Mastering Algebra - Advanced Level 2.4 One-to-One and Inverse Functions of Real Variables 
Hamilton Education Guides 146 
; 
2
12 +
=
xy = ( )
2
121 +=− xxf 
Second Method: Since ( )[ ]xff 1− = x replace x with ( )xf 1− wherever x is present in the 
function ( ) 12 += xxf , i.e., ( )[ ] ( ) 12 11 −= −− xfxff . 
Next, solve for ( )xf 1− by equating the right hand side of the equations ( )[ ] xxff =−1 and 
( )[ ] ( ) 12 11 −= −− xfxff , i.e., ( ) 12 1 −= − xfx ; ( )( ) 2112 1 −= − xfx ; ( )( ) 212 2112 ×− −= xfx 
 
; ( ) 12 12 −= − xfx ; ( )xfx 12 21 −=+ ; ( )
2
2
2
1 12 xfx −
=
+ ; ( )xfx 1
2
2
1 −=+ ; ( )
2
121 +=− xxf 
f. The function ( ) 55 −= xxf is a one-to-one function because for each value of 0≥x there is only 
 one corresponding value of y , i.e., at 0=x ; 5−=y or at 52=x ; 3−=y , etc. 
First Method: Interchange the x variable with y and the y variable with x in the equation 
( ) 55 −== xyxf and solve for y . Next, replace y with ( )xf 1− to obtain: 
55 −= yx ; 55 yx =+ ; 5
1
5 yx =+ ; ( ) 55 5
1
5
×
=+ yx ; ( ) yx =+ 55 ; ( )55+= xy ; ( ) ( )51 5+=− xxf 
Second Method: Since ( )[ ]xff 1− = x replace x with ( )xf 1− wherever x is present in the 
function ( ) 55 −= xxf , i.e., ( )[ ] ( ) 55 11 −= −− xfxff . 
Next, solve for ( )xf 1− by equating the right hand side of the equations ( )[ ] xxff =−1 and 
( )[ ] ( ) 55 11 −= −− xfxff , i.e., ( ) 55 1 −= − xfx ; ( )5 15 xfx −=+ ; ( )[ ] 5115 xfx −=+ ; ( ) ( )[ ] 515 515 ×−=+ xfx 
 
; ( ) ( )xfx 155 −=+ ; ( ) ( )51 5+=− xxf 
g. The function ( ) 53 5 −= xxf is a one-to-one function because for each value of x there is only 
one corresponding value of y . For example, at 0=x ; 5−=y or at 1=x ; 2−=y , etc. 
First Method: Interchange the x variable with y and the y variable with x in the equation 
( ) 53 5 −== xyxf and solve for y . Next, replace y with ( )xf 1− to obtain: 
53 5 −= yx ; 535 yx =+ ; 
3
3
3
5 5yx
=
+ ; 5
3
5 yx =+ ; 
5
1
5
1
3
55





 +=
× xy ; 5
3
5+
=
xy ; ( ) 51
3
5+
=−
xxf 
Second Method: Since ( )[ ]xff 1− = x replace x with ( )xf 1− wherever x is present in the 
function ( ) 53 5 −= xxf , i.e., ( )[ ] ( )[ ] 53 511 −⋅= −− xfxff . 
Next, solve for ( )xf 1− by equating the right hand side of the equations ( )[ ] xxff =−1 and 
 
Mastering Algebra - Advanced Level 2.4 One-to-One and Inverse Functions of Real Variables 
Hamilton Education Guides 147 
( )[ ] ( )[ ] 53 511 −⋅= −− xfxff , i.e., ( )[ ] 53 51 −⋅= − xfx ; ( )[ ]5135 xfx −⋅=+ ; ( )[ ]
3
3
3
5
51 xfx
−⋅
=
+ ; 
3
5+x 
 
= ( )[ ] 51 xf − ; ( )[ ] 515
1
51
3
5 ×−=




 + xfx ; ( )xfx 15
1
3
5 −=




 + ; ( ) 51
3
5+
=−
xxf 
h. The function ( ) 5
3
1
−= xxf is a one-to-one function because for each value of x there is only 
one corresponding value of y . For example, at 0=x ; 5−=y or at 3=x ; 4−=y , etc. 
First Method: Interchange the x variable with y and the y variable with x in the equation 
( ) 5
3
1
−== xyxf and solve for y . Next, replace y with ( )xf 1− to obtain: 
5
3
1
−= yx ; yx
3
15 =+ ; ( ) yx
3
1353 ×=+× ; yx =+153 ; 153 += xy ; ( ) 1531 +=− xxf 
Second Method: Since ( )[ ]xff 1− = x replace x with ( )xf 1− wherever x is present in the 
function ( ) 5
3
1
−= xxf , i.e., ( )[ ] ( ) 5
3
1 11 −⋅= −− xfxff . 
Next, solve for ( )xf 1− by equating the right hand side of the equations ( )[ ] xxff =−1 and 
( )[ ] ( ) 5
3
1 11 −⋅= −− xfxff , i.e., ( ) 5
3
1 1 −⋅= − xfx ; ( )
3
5
1 xfx
−
=+ ; ( ) ( )xfx 153 −=+ ; ( ) 1531 +=− xxf 
i. The function ( ) 100+= xxf isa one-to-one function because for each value of x there is only 
 one corresponding value of y , i.e., at 0=x ; 100=y or at 10=x ; 110=y , etc. 
First Method: Interchange the x variable with y and the y variable with x in the equation 
( ) 100+== xyxf and solve for y . Next, replace y with ( )xf 1− to obtain: 
100+= yx ; yx =−100 ; 100−= xy ; ( ) 1001 −=− xxf 
Second Method: Since ( )[ ]xff 1− = x replace x with ( )xf 1− wherever x is present in the 
function ( ) 100+= xxf , i.e., ( )[ ] ( ) 10011 += −− xfxff . 
Next, solve for ( )xf 1− by equating the right hand side of the equations ( )[ ] xxff =−1 and 
( )[ ] ( ) 10011 += −− xfxff , i.e., ( ) 1001 += − xfx ; ( )xfx 1100 −=− ; ( ) 1001 −=− xxf 
j. The function ( ) xxf 4.0= is a one-to-one function because for each value of x there is only one 
 corresponding value of y . For example, at 0=x ; 0=y or at 2=x ; 8.0=y , etc. 
First Method: Interchange the x variable with y and the y variable with x in the equation 
( ) xyxf 4.0== and solve for y . Next, replace y with ( )xf 1− to obtain: 
yx 4.0= ; 
4.0
4.0
4.0
yx
= ; yx =
4.0
 ; xy 5.2= ; ( ) xxf 5.21 =− 
Mastering Algebra - Advanced Level 2.4 One-to-One and Inverse Functions of Real Variables 
Hamilton Education Guides 148 
Second Method: Since ( )[ ]xff 1− = x replace x with ( )xf 1− wherever x is present in the 
function ( ) xxf 4.0= , i.e., ( )[ ] ( )xfxff 11 4.0 −− ⋅= . 
Next, solve for ( )xf 1− by equating the right hand side of the equations ( )[ ] xxff =−1 and 
( )[ ] ( )xfxff 11 4.0 −− ⋅= , i.e., ( )xfx 14.0 −⋅= ; ( )
4.0
4.0
4.0
1 xfx −
= ; ( )xfx 15.2 −= ; ( ) xxf 5.21 =− 
k. The function ( ) 31 +=
x
xf is a one-to-one function because for each value of x (except at 0=x ) 
there is only one corresponding value of y . For example, at 1=x ; 4=y or at 2=x ; 5.3=y , etc. 
First Method: Interchange the x variable with y and the y variable with x in the equation 
( ) 31 +==
x
yxf and solve for y . Replace y with ( )xf 1− to obtain: 
31 +=
y
x ; 
y
x 13 =− ; ( ) 13 =−xy ; 
3
1
−
=
x
y ; ( )
3
11
−
=−
x
xf 
Second Method: Since ( )[ ]xff 1− = x replace x with ( )xf 1− wherever x is present in the 
function ( ) 31 +=
x
xf , i.e., ( )[ ]
( )
31
1
1 +=
−
−
xf
xff . 
Next, solve for ( )xf 1− by equating the right hand side of the equations ( )[ ] xxff =−1 and 
( )[ ]
( )
31
1
1 +=
−
−
xf
xff , i.e., 
( )
31
1
+=
− xf
x ; 
( )xf
x
1
13
−
=− ; ( ) ( ) 13 1 =⋅− − xfx ; ( )
3
11
−
=−
x
xf 
l. The function ( ) xxf −= is a one-to-one function because for each value of x there is only one 
corresponding value of y . For example, at 0=x ; 0=y or at 1=x ; 1−=y , etc. 
First Method: Interchange the x variable with y and the y variable with x in the equation 
( ) xyxf −== and solve for y . Replace y with ( )xf 1− to obtain: yx −= ; xy −= ; ( ) xxf −=−1 
Second Method: Since ( )[ ]xff 1− = x replace x with ( )xf 1− wherever x is present in the 
function ( ) xxf −= , i.e., ( )[ ] ( )xfxff 11 −− −= . 
Next, solve for ( )xf 1− by equating the right hand side of the equations ( )[ ] xxff =−1 and 
( )[ ] ( )xfxff 11 −− −= , i.e., ( )xfx 1−−= ; ( ) xxf −=−1 
m. The function ( ) 375 +−= xxf is a one-to-one function because for each value of 7≥x there is 
only one corresponding value of y , i.e., at 7=x ; 3=y or at 10=x ; 25.4335 =+=y , etc. 
First Method: Interchange the x variable with y and the y variable with x in the equation 
( ) 375 +−== xyxf and solve for y . Next, replace y with ( )xf 1− to obtain: 
375 +−= yx ; 5 73 −=− yx ; ( )5173 −=− yx ; ( ) ( ) 55 5173 ×−=− yx ; ( ) 73 5 −=− yx ; ( ) yx =+− 73 5 
Mastering Algebra - Advanced Level 2.4 One-to-One and Inverse Functions of Real Variables 
Hamilton Education Guides 149 
= ( ) 73 5 +−= xy = ( ) ( ) 73 51 +−=− xxf 
Second Method: Since ( )[ ]xff 1− = x replace x with ( )xf 1− wherever x is present in the 
function ( ) 375 +−= xxf , i.e., ( )[ ] ( ) 375 11 +−= −− xfxff . 
Next, solve for ( )xf 1− by equating the right hand side of the equations ( )[ ] xxff =−1 and 
( )[ ] ( ) 375 11 +−= −− xfxff , i.e., ( ) 375 1 +−= − xfx ; ( )5 1 73 −=− − xfx ; ( )[ ] 5173 1 −=− − xfx 
 
; ( ) ( )[ ] 515 5173 ×− −=− xfx ; ( ) ( ) 73 15 −=− − xfx ; ( ) ( )xfx 15 73 −=+− ; ( ) ( ) 73 51 +−=− xxf 
n. The function ( )
1
52
−
−
=
x
xxf is a one-to-one function because for each value of x (except at 1=x ) 
there is only one corresponding value of y . For example, at 2=x ; 1−=y or at 3=x ; 5.0=y , etc. 
First Method: Interchange the x variable with y and the y variable with x in the equation 
( )
1
52
−
−
==
x
xyxf and solve for y . Replace y with ( )xf 1− to obtain: 
1
52
−
−
=
y
yx ; ( ) 521 −=− yyx ; 52 −=− yxxy ; 52 −=− xyxy ; ( ) 52 −=− xxy ; 
2
5
−
−
=
x
xy ; ( )
2
51
−
−
=−
x
xxf 
Second Method: Since ( )[ ]xff 1− = x replace x with ( )xf 1− wherever x is present in the 
function ( )
1
52
−
−
=
x
xxf , i.e., ( )[ ] ( )
( ) 1
52
1
1
1
−
−
=
−
−
−
xf
xfxff . 
Next, solve for ( )xf 1− by equating the right hand side of the equations ( )[ ] xxff =−1 and 
( )[ ] ( )
( ) 1
52
1
1
1
−
−
=
−
−
−
xf
xfxff , i.e., ( )
( ) 1
52
1
1
−
−
=
−
−
xf
xfx ; ( )[ ] ( ) 521 11 −=−⋅ −− xfxfx ; ( ) ( ) 52 11 −=− −− xfxxfx 
 
( ) ( ) 52 11 −=− −− xxfxfx ; ( ) ( ) 521 −=−⋅− xxxf ; ( )
2
51
−
−
=−
x
xxf 
o. The function ( ) xxf 37 −= is a one-to-one function because for each value of x there is only 
one corresponding value of y . For example, at 0=x ; 7=y or at 
3
1
=x ; 6=y , etc. 
First Method: Interchange the x variable with y and the y variable with x in the equation 
( ) xyxf 37 −== and solve for y . Next, replace y with ( )xf 1− to obtain: 
yx 37 −= ; yx 37 −=− ; 
3
3
3
7
−
−
=
−
− yx ; yx =−
3
7 ; 
3
7 xy −= ; ( )
3
71 xxf −=− 
Second Method: Since ( )[ ]xff 1− = x replace x with ( )xf 1− wherever x is present in the 
function ( ) xxf 37 −= , i.e., ( )[ ] ( )xfxff 11 37 −− ⋅−= . 
Next, solve for ( )xf 1− by equating the right hand side of the equations ( )[ ] xxff =−1 and 
Mastering Algebra - Advanced Level 2.4 One-to-One and Inverse Functions of Real Variables 
Hamilton Education Guides 150 
( )[ ] ( )xfxff 11 37 −− ⋅−= , i.e., ( )xfx 137 −⋅−= ; ( )xfx 137 −−=− ; ( )
3
3
3
7 1
−
−
=
−
− − xfx ; ( )
3
71 xxf −=− 
Note that by limiting the domain of a function, a non-invertible function can become an 
invertible function. For example, the function ( ) 2xxf = is not invertible. However, by defining 
( ) 2xxf = where 0≥x the function ( ) 2xxf = becomes an invertible function. The following are 
additional examples of non-invertible functions where by restricting the domain result in having 
invertible functions: 
( ) 32 += xxf where 0≥x ( ) 2xxf = where 0≤x 
( ) 31 +−= xxf where 1≥x ( ) 4
2
1
+−= xxf where 0≤x 
( ) xxf
2
1
= where 0≥x ( ) ( ) 432 2 −+= xxf where 3−≥x 
( ) ( )22−= xxf where 2≥x ( ) 21 xxf −= where 0≤x 
In the next section we will learn how to write complex numbers in standard form and perform 
math operations involving complex numbers. 
Section 2.4 Practice Problems – One-to-One and Inverse Functions of Real Variables 
1. State which of the following functions are one-to-one. 
a. ( ) 52 += xxf b. ( ) xxf
3
25+−= c. ( ) 1
5
1
−= xxf 
d. ( ) 252 −= xxf e. ( ) 56 −= xxf f. ( ) ( )xxf 316
4
1
−= 
g. ( ) 23 −= xxf h. ( ) xxf 2= i. ( ) 4xxf = 
j. ( ) xexf 21+= k. ( ) 18 += xxf l. ( ) 42 += xxf 
2. Given the following functions are one-to-one, use the first method to find their inverse. 
a. ( ) 3+= xxf b. ( ) xxf 5= c. ( ) 15 −= xxf 
d. ( ) 321 xxf −= e. ( ) 12 += xxf , 0≥x f. ( ) 102.0 += xxf 
g.( ) 123 +−= xxf , 2≥x h. ( )
x
xxf 32 −= , 0≠x i. ( ) xxf 52−= 
j. ( )
1
52
+
+
=
x
xxf , 1−≠x k. ( )
5
32
−
−
=
x
xxf , 5≠x l. ( ) 92 3 −= xxf 
Mastering Algebra - Advanced Level 2.5 Complex Numbers and Functions of Complex Variables 
Hamilton Education Guides 151 
2.5 Complex Numbers and Functions of Complex Variables 
In Chapters 2 and 4 of the “Mastering Algebra – Intermediate Level ” we solved equations of the 
form 0=+ bax and 02 =++ cbxax and learned that: 
1. Any linear equation such as 423 −=+x and 0153 =−x has one real solution ( )2−=x and ( )5=x , 
respectively and 
2. Any quadratic equation such as 062 =−− xx and 162 =x has two real solutions ( )23 −and and 
( )44 −and , respectively. 
However, quadratic equations of the form 52 −=x have no real solutions because the square of 
every real number is positive. Therefore, in order for equations such as 52 −=x to have a 
solution imaginary numbers are introduced. Imaginary numbers are generally shown by the 
letters “ i ” or “ j ”. Imaginary numbers are combined with real numbers to form complex 
numbers. Numbers of the form bia + , where a and b are real numbers, are defined as complex 
numbers. Note that in the complex number bia + , a is referred to as the real part and b is 
referred to as the imaginary part. The imaginary number i is defined as 
i = 1− and 2i = ( )21− = 2211 ×− = 1− 
The form bia + is referred to as the standard form of a complex number. Furthermore, two 
complex numbers bia + and dic + are equal to one another if and only if ca = and db = . The 
following examples show how higher order imaginary numbers are simplified: 
Example 2.5-1: Simplify the following imaginary numbers. 
a. 3i = b. 4i = c. 5i = d. 8i = 
e. 7i = f. 14i = g. 15i = h. 20i = 
i. 18i = j. 22i = k. 25i = l. 40i = 
Solutions: 
a. 3i = 12+i = ii ⋅2 = i⋅−1 = i− b. 4i = ( )22i = ( )21− = 1 
 
c. 5i = 14+i = ii ⋅4 = ( ) ii ⋅22 = ( ) i⋅− 21 = i⋅1 = i d. 8i = ( )42i = ( )41− = 1 
 
e. 7i = 16+i = 16 ii ⋅ = ( ) ii ⋅32 = ( ) i⋅− 31 = i⋅−1 = i− f. 14i = ( )72i = ( )71− = 1− 
 
g. 15i = 114+i = ii ⋅14 = ( ) ii ⋅72 = ( ) i⋅− 71 = i⋅−1 = i− h. 20i = ( )102i = ( )101− = 1 
 
i. 18i = ( )92i = ( )91− = 1− j. 22i = ( )112i = ( )111− = 1− 
 
k. 25i = 124+i = ii ⋅24 = ( ) ii ⋅122 = ( ) i⋅− 121 = i⋅1 = i l. 40i = ( )202i = ( )201− = 1 
 
Mastering Algebra - Advanced Level 2.5 Complex Numbers and Functions of Complex Variables 
Hamilton Education Guides 152 
There are many ways to simplify higher powers of i . However, to simplify the higher powers of 
i , it is much easier to use 12 −=i and the exponent laws as shown in the above examples (review 
Section 1.1). Note that all powers of i can be simplified to 1 , 1− , 1−=i , or 1−−=− i . 
Note 1: The number “negative one” raised to an even number is always equal to plus one. For 
Example: ( ) ( ) ( ) ( ) ( ) ( ) ( ) .,1,1,1,1,1,1,1 1005624181082 etc−−−−−−− are all equal to 1 . 
Note 2: The number “negative one” raised to an odd number is always equal to negative one. For 
Example: ( ) ( ) ( ) ( ) ( ) ( ) ( ) .,1,1,1,1,1,1,1 1014523191373 etc−−−−−−− are all equal to 1− . 
Example 2.5-2: Determine whether the following complex numbers are real or imaginary. 
Write the complex number in the standard form bia + . 
a. i5 = b. 52 +− i = c. 
5
21 i+ = d. 
2
3π = 
e. i31− = f. 7 = g. i− = h. 82 i+ = 
Solutions: 
a. The complex number i5 is imaginary. The standard form of i5 is i50 + . 
b. The complex number 52 +− i is imaginary. The standard form of 52 +− i is i25 − . 
c. The complex number 
5
21 i+ is imaginary. The standard form of 
5
21 i+ is i
5
2
5
1
+ = i4.02.0 + . 
d. The complex number 
2
3π is real (an irrational number). The standard form of 
2
3π is i0
2
3
+
π . 
e. The complex number i31− is imaginary. The standard form of i31− is ( )i31 −+ . 
f. The complex number 7 is real (a real number). The standard form of 7 is i07 + . 
g. The complex number i− is imaginary. The standard form of i− is ( )i−+0 . 
h. The complex number 82 i+ = ( )422 i+ = ( )412 −+ = 12+ = 3 is real (a real number). The 
standard form of 3 is i03 + . 
Note 3: In the real number system, 6− , 9− , and 23− are undefined. However, in the 
complex number system they are defined as i66 =− , i99 =− , and i2323 =− . In 
addition, note that all math operations involving complex numbers must be performed after 
converting to the standard form bia + . For example, to add 5− to 125− we must first 
replace the square roots of the negative numbers by square roots represented by i , i.e., 
i5155 =−⋅=− and 125 i1251125 =−⋅= before performing the addition. The 
following examples further illustrate this point. 
Example 2.5-3: Write the following expressions in the standard form bia + . 
a. 1255 −+− = b. ( ) 9273 −−−+− = c. 
4
24 −−− = 
d. 169 −+− = e. 52 −−− = f. ( )25− = 
Mastering Algebra - Advanced Level 2.5 Complex Numbers and Functions of Complex Variables 
Hamilton Education Guides 153 
g. ( )33− = h. 425 −− = i. ( )924 −+−− = 
j. 
3
23 −+− = k. 25 −⋅− = l. ( )2364 −−− = 
m. 
2
43
−
−− = n. ( )1653 −−−− = o. ( ) 513 2 −−+ = 
Solutions: 
a. 1255 −+− = ( ) ( )112515 −⋅+−⋅ = ( ) ( )ii ⋅+⋅ 1255 = ii 555 + = ii 18.1124.2 + = i42.130+ 
 
b. ( ) 9273 −−−+− = ( ) ( )1912713 −⋅−−⋅+−⋅ = iii 9273 −+ = iii 32.573.1 −+ 
 
 = i93.3 = i93.30 + 
 
c. 
4
24 −−− = ( ) ( )
4
1214 −⋅−−⋅ = 
4
24 ii − = 
4
41.12 ii − = i
4
59.0 = i15.0 = i15.00 + 
 
d. 169 −+− = ( ) ( )11619 −⋅+−⋅ = ( ) ( )ii ⋅+⋅ 43 = ii 43 + = i7 = i70+ 
 
e. 52 −−− = ( ) ( )1512 −⋅−−⋅ = ( ) ( )ii ⋅−⋅ 24.241.1 = ii 24.241.1 − = i83.0− = ( )i83.00 −+ 
 
f. ( )25− = ( )215 −⋅ = ( )25 i⋅ = 25i = 5− = i05 +− 
 
g. ( )33− = ( )313 −⋅ = ( )373.1 i⋅ = 318.5 i⋅ = i−⋅18.5 = i18.5− = ( )i18.50 −+ 
 
h. 425 −− = ( ) 2125 −−⋅ = ( ) 25 −⋅ i = 25 −i = i52 +− 
 
i. ( )924 −+−− = ( ) ( ) ( )[ ]191214 −⋅+−⋅−⋅ = ( ) ( ) ( )[ ]iii ⋅+⋅⋅ 924 = ( )iii 341.12 + 
 
= 22 682.2 ii + = 682.2 −− = 82.8− = i082.8 +− 
 
j. 
3
23 −+− = 
3
1213 −⋅+−⋅ = 
3
23 ii ⋅+⋅ = 
3
41.173.1 ii + = 
3
14.3 i = i05.1 = i05.10 + 
 
k. 25 −⋅− = ( ) ( )1215 −⋅⋅−⋅ = ii 25 ⋅ = ( ) ( )ii ⋅⋅⋅ 41.124.2 = 216.3 i = 16.3− = i016.3 +− 
 
Again note that math operations involving complex numbers must be performed after converting 
to the standard form bia + , i.e., in the above example we must first convert 5− and 2− to i5 
and i2 before multiplying the two number by one another. (We can not multiply 5− by 2− 
Mastering Algebra - Advanced Level 2.5 Complex Numbers and Functions of Complex Variables 
Hamilton Education Guides 154 
to obtain i016.316.31025 +===−⋅− . This is the wrong answer. The product rule 
abba =⋅ can be used only when a and b are positive numbers.) 
 
l. ( )2364 −−− = ( )213614 −⋅−−⋅ = ( )262 ii − = ( )24i− = 216i = 16− = i016 +− 
 
m. 
2
43
−
−− = 
2
143
−
−⋅− = 
2
43
−
⋅− i = 
2
23
−
− i = 
2
2
2
3
−
−
+
−
i = i
2
2
2
3
−
−
+− = i+− 5.1 
 
n. ( )1653 −−−− = ( ) ( ) ( )[ ]1161513 −⋅−−⋅−⋅ = ( ) ( )[ ]iii ⋅−⋅⋅ 1653 = ( )iii 453 − 
 
= ( )iii 424.273.1 − = ii 76.173.1 −× = 204.3 i = 04.3− = i004.3 +− 
 
o. ( ) 513 2 −−+ = 513 −+ = 43 −+ = 143 −⋅+ = i43+ = i23 + 
In Chapter 4, Section 4.2, of the “Mastering Algebra – Intermediate Level” we learned that one 
of the methods for solving quadratic equations of the form 02 =++ cbxax (where a , b , and c are 
real numbers and 0≠a ) is by using the quadratic formula defined as 
a
acbbx
2
42
2,1
−±−
= . 
However, to find the solutions the term under the radical,i.e., acb 42 − had to be real and 
positive. The introduction of complex numbers enables us to solve quadratic equations even 
though the term under the radical, i.e., acb 42 − is negative. The following examples illustrate 
this point. 
Example 2.5-4: Given the quadratic formula 
a
acbbx
2
42
2,1
−±−
= , solve for x for the following 
values of a , b , and c . 
a. 1=a , 2=b , and 6=c b. 2=a , 0=b , and 4=c 
c. 1=a , 1−=b , and 3=c d. 3=a , 2=b , and 5=c 
Solutions: 
a. Substituting 1=a , 2=b , and 6=c into 
a
acbbx
2
42
2,1
−±−
= we obtain 
1x = 
( )
12
61422 2
⋅
⋅⋅−+− = 
2
2442 −+− = 
2
202 −+− = 
2
1202 −⋅+− = 
2
542 i⋅⋅+− = 
2
522 i+− 
 
 = ( )
2
512
/
+−/ i = i51+− = i24.21+− and 
 
2x = 
( )
12
61422 2
⋅
⋅⋅−−− = 
2
2442 −−− = 
2
202 −−− = 
2
1202 −⋅−− = 
2
542 i⋅⋅−− = 
2
522 i−− 
 
Mastering Algebra - Advanced Level 2.5 Complex Numbers and Functions of Complex Variables 
Hamilton Education Guides 155 
 = ( )
2
512
/
+/− i = i51−− = i24.21−− 
b. Substituting 2=a , 0=b , and 4=c into 
a
acbbx
2
42
2,1
−±−
= we obtain 
1x = 
( )
22
42400 2
⋅
⋅⋅−+− = 
4
320 −+− = 
4
132 −⋅ = 
4
216 i⋅⋅ = 
4
24
/
/ i = i2 = i41.10+ and 
 
2x = 
( )
22
42400 2
⋅
⋅⋅−−− = 
4
320 −−− = 
4
132 −⋅− = 
4
216 i⋅⋅− = 
4
24
/
/
−
i = i2− = i41.10− 
c. Substituting . 1=a , 1−=b , and 3=c into 
a
acbbx
2
42
2,1
−±−
= we obtain 
1x = 
( ) ( ) ( )
12
31411 2
⋅
⋅⋅−−+−− = 
2
1211 −+ = 
2
111 −+ = 
2
1111 −⋅+ = 
2
111 i⋅+ = 
2
32.31 i+ 
 
 = i
2
32.3
2
1
+ = i66.15.0 + and 
 
2x = 
( ) ( ) ( )
12
31411 2
⋅
⋅⋅−−−−− = 
2
1211 −− = 
2
111 −− = 
2
1111 −⋅− = 
2
111 i⋅− = 
2
32.31 i− 
 
 = i
2
32.3
2
1
− = i66.15.0 − 
d. Substituting 3=a , 2=b , and 5=c into 
a
acbbx
2
42
2,1
−±−
= we obtain 
1x = 
( )
32
53422 2
⋅
⋅⋅−+− = 
6
6042 −+− = 
6
562 −+− = 
6
1562 −⋅+− = 
6
1442 i⋅⋅+− 
 
 = 
6
1422 i+− = ( )
3
6
1412
/
+−/ i = 
3
74.31 i+− = i
3
74.3
3
1
+− = i25.133.0 +− and 
 
2x = 
( )
32
53422 2
⋅
⋅⋅−−− = 
6
6042 −−− = 
6
562 −−− = 
6
1562 −⋅−− = 
6
1442 i⋅⋅−− 
 
 = 
6
1422 i−− = ( )
3
6
1412
/
−−/ i = 
3
74.31 i−− = i
3
74.3
3
1
−− = i25.133.0 −− 
Note that the solutions to the quadratic equations, for cases where the radicand is negative, are 
complex conjugates of one another. 
Mastering Algebra - Advanced Level 2.5 Complex Numbers and Numbers of Complex Variables 
Hamilton Education Guides 156 
Similar to real numbers (see Section 2.1), complex numbers can be substituted in place of a 
variable in functions of one variable as shown in the following examples: 
Example 2.5-5: Given the functions ( ) 322 ++= xxxf and ( ) 72 += xxg , find. 
a. ( )if +−1 = b. ( )ig −2 = c. ( )if +1 = d. ( )if 2 = 
e. ( )ig 31− = f. ( )if 43+ = g. ( )ig − = h. ( )if = 
Solutions: 
a. ( )if +−1 = ( ) ( ) 3121 2 ++−++− ii = 32221 2 ++−−+ iii = 322211 +//+−//−/−/ ii = 23− = 1 = i01 + 
 
b. ( )ig −2 = ( ) 722 +− i = 724 +− i = ( ) i274 −+ = i211− 
 
c. ( )if +1 = ( ) ( ) 3121 2 ++++ ii = 32221 2 +++++ iii = 322211 ++++/−/ ii = ( ) ( )ii 2232 +++ = i45 + 
 
d. ( )if 2 = ( ) ( ) 3222 2 ++ ii = 3222 2 ++ ii = 3222 ++− i = ( ) i83.223 +− = i83.21+ 
 
e. ( )ig 31− = ( ) 7312 +− i = 7322 +− i = ( ) i3272 −+ = i46.39 − 
 
f. ( )if 43+ = ( ) ( ) 343243 2 ++++ ii = 38624169 2 +++++ iii = 38624169 ++++− ii = ( )36169 ++− 
 
( )ii 824 ++ = i322 + 
 
g. ( )ig − = ( ) 72 +−i = 72 +− i = i27 − 
 
h. ( )if = 322 ++ ii = 321 ++− i = ( ) i213 +− = i22 + 
Again, similar to real numbers (see Section 2.3), complex numbers can also be substituted in 
place of a variable in composite functions as shown in the following examples. 
Example 2.5-6: Given ( ) 5+= xxf and ( ) 122 +−= xxxg , find 
a. ( )( )igf 32+ = b. ( )( )ifg +1 = c. ( )( )igf −1 = 
d. ( )( )ifg = e. ( )( )igf 23 + = f. ( )( )32 ifg + = 
g. ( )( )iifg +2 = h. ( )( )5igf = i. ( )( )71 igf − = 
Solutions: 
First - Find ( )( )xgf and ( )( )xfg , i.e., 
( )( )xgf = ( ) 5+xg = ( ) 5122 ++− xx = 622 +− xx 
 
( )( )xfg = ( )[ ] ( ) 122 +⋅− xfxf = ( ) ( ) 1525 2 ++−+ xx = 110225102 +−−++ xxx = 1682 ++ xx 
 
 
 
Mastering Algebra - Advanced Level 2.5 Complex Numbers and Numbers of Complex Variables 
Hamilton Education Guides 157 
Second – Find ( )( )xgf or ( )( )xfg for the specific values given. 
a. ( )( )igf 32+ = ( ) ( ) 632232 2 ++−+ ii = 6641294 2 +−−++ iii = ( ) ( )ii 6126494 −++−− = i63 +− 
 
b. ( )( )ifg +1 = ( ) ( ) 16181 2 ++++ ii = 168821 2 +++++ iii = 1688211 ++++/−/ ii = i1024 + 
 
c. ( )( )igf −1 = ( ) ( ) 6121 2 +−−− ii = 62221 2 ++−−+ iii = 622211 +//+−//−/−/ ii = 4 = i04 + 
 
d. ( )( )ifg = 1682 ++ ii = 1681 ++− i = ( ) i8116 +− = i815+ 
 
e. ( )( )igf 23 + = ( ) ( ) 623223 2 ++−+ ii = 64323443 2 +−−++ iii = 6446.393.643 +−−+− ii 
 
= ( ) ( )ii 493.6646.343 −++−− = i93.254.1 + 
 
f. ( )( )32 ifg + = ( )( )122 ++ ifg = ( )( )iifg ⋅+ 22 = ( )( )ifg −2 = ( ) ( ) 16282 2 +−+− ii = 1681644 2 +−+−+ iii 
 
= 16816414 +−+−− ii = ( ) ( )ii 84161614 −−+++− = i1235− 
 
g. ( )( )iifg +2 = ( )( )ifg +−1 = ( ) ( ) 16181 2 ++−++− ii = 168821 2 ++−−+ iii = 1688211 ++−−/−/ ii 
 
= ( ) ( )ii 82168 +−++− = i68 + 
 
h. ( )( )5igf = ( )( )14+igf = ( )( )iigf ⋅4 = ( )( )igf = 622 +− ii = 621 +−− i = ( ) i261 −+− = i25 − 
 
i. ( )( )71 igf − = ( )( )161 +− igf = ( )( )iigf ⋅− 61 = ( )( )igf +1 = ( ) ( ) 6121 2 ++−+ ii = 62221 2 +−−++ iii 
 
= 6211 +−− = 4 = i04 + 
Example 2.5-7: Given ( )( )( ) 1022 ++= xxxhgf and ( )( )( ) 53 += xxhfg , find 
a. ( )( )( )21 −−hgf = b. ( )( )( )ihgf − = c. ( )( )( )3ihfg = 
d. ( )( )( )4−hfg = e. ( )( )( )31 ihgf − = f. ( )( )( )ihfg 53+ = 
Solutions: 
a. ( )( )( )21 −−hgf = ( )( )( )121 −⋅−hgf = ( )( )( )ihgf 21− = ( ) ( ) 1021221 2 +−+− ii 
 
= 102222221 2 +−+−+ iii = 102222221 +−+−− ii = ( ) ( )ii 222210221 −−+++− 
 
= i2411− = ( )i414.1411 ⋅− = i66.511− 
 
 
 
 
 
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Hamilton Education Guides 158 
b. ( )( )( )ihgf − = 1022 ++ ii = 1021 ++− i = ( ) i2101 ++− = i29 + 
 
c. ( )( )( )3ihfg = ( )( )( )12+ihfg = ( )( )( )iihfg ⋅2 = ( )( )( )ihfg − = ( ) 53 +−⋅ i = 53 +− i = i35 − 
 
d. ( )( )( )4−hfg = ( )( )( )ihfg 4 = ( )( )( )ihfg 2 = ( ) 523 +⋅ i = 56 +i = i65 + 
 
e. ( )( )( )31 ihgf − = ( )( )( )ihgf +1 = ( ) ( ) 10121 2 ++++ ii = 102221 2 +++++ iii = 1022211 ++++− ii 
 
= 10222 +++ ii = ( ) ( )ii 22102 +++ = i412 + 
 
f. ( )( )( )ihfg 53+ = ( ) 5533 ++ i = 5159 ++ i = ( ) i1559 ++ = i1514+ 
 
In the following section students are introduced to operations involving the addition, subtraction, 
multiplication, and division of complex numbers. 
Practice Problems – Complex Numbers and Functions of Complex Variables 
1. Simplify the following imaginary numbers. 
 a. 10i = b. 13i = c. 17i = d. 21i = 
 e. 50i = f. 100i = g. 47i = h. 29i = 
2. Write the following expressions in the standard form bia + . 
 a. 126 −+− = b. ( )432 −⋅− = c. 53 −⋅− = 
 d. ( )3252 −−− = e. ( )321 −+−− = f. ( ) 215 3 −−− = 
3. Given ( ) 12 += xxf and ( ) 122 +−= xxxg , find 
 a. ( )if −1 = b. ( )if − = c. ( )ig 21− = 
 d. ( )31 ig + = e. ( )if +1 = f. ( )if 32+ = 
 ( )1−−g = h. ( )ig +2 = i. ( )33 −+f = 
4. Given ( ) 12 −= xxf and ( ) 5−= xxg , find 
 a. ( )( )11 −−gf = b. ( )( )ifg −1 = c. ( )( )4ifg − = 
 d. ( )( )igf 52+ = e. ( )( )6igf = f. ( )( )51 ifg + = 
5. Given the quadratic formula 
a
acbbx
2
42
2,1
−±−
= ,solve for x for the following values of a , 
 b , and c . 
 a. 2=a , 3=b , and 5=c b. 3=a , 4=b , and 6=c c. 4=a , 1=b , and 10=c 
 
Mastering Algebra - Advanced Level 2.6 Math Operations Involving Complex Numbers 
Hamilton Education Guides 159 
2.6 Math Operations Involving Complex Numbers 
In this section addition, subtraction, multiplication, division, and mixed operations involving 
complex numbers are addressed in Cases I through IV. Several examples showing the math 
operations with complex numbers are presented. Students are encouraged to learn how to solve 
the following cases involving complex numbers for future use in calculus. 
Case I Addition and Subtraction of Complex Numbers 
In general, complex numbers are added and subtracted by grouping the real and imaginary parts 
together as shown below. 
( ) ( )idciba +++ = ( ) ( ) idbca +++ 
( ) ( )idciba +−+ = ( ) ( ) idbca −+− 
Note that the commutative and associative properties for addition and subtraction of real numbers 
are also valid for complex numbers as long as i is treated as a variable. In addition, note that 
addition and subtraction of complex numbers is similar to addition and subtraction of like terms in 
exponents (see Section 1.1b, Case III) 
Example 2.6-1 
( ) ( )ii 4523 −++ = ( ) ( )ii 4253 −++ = ( ) ( )i4253 −++ = i28− 
Example 2.6-2 
( ) ( )ii 6343 −−−− = ( ) ( )ii 6343 +−+−− = ( ) ( )ii 6433 +−+−− = ( )i646 +−+− = i26 +− 
Example 2.6-3 
( ) ii 285 −− = ii 285 −− = ( )i285 −−+ = i105 − 
Example 2.6-4 
( ) ( )ii 5224 +−−− = ( ) ( )ii 5224 −+− = ( ) ( )ii 5224 −−−+ = ( )i526 −−+ = i76 − 
Example 2.6-5 
( ) ( ) ( )[ ]iii 825132 −−−++ = ( ) ( ) ( )[ ]iii 852132 +−+−++ = ( ) ( )ii 3132 +−++ = ( ) ( )ii 3312 ++− = i61 + 
Example 2.6-6 
( )[ ] ( )iii 326.338.5 −+−+ = ( )[ ] ( )ii 326.38.53 −+−+ = ( ) ( )ii 322.23 −++ = ( ) ( )ii 32.223 −++ = i8.05− 
Example 2.6-7 
( )[ ] ( )iii 5.141.25.33 −−++ = ( )[ ] ( )iii 5.141.235.3 +−+++ = ( ) ( )ii 5.141.55.3 +−++ = ( ) ( )ii 5.11.545.3 ++− 
 
= ( ) ( )i5.11.545.3 ++− = i6.65.0 +− 
Example 2.6-8 
( ) ( )93452 −−−−+ = ( ) ( )1331252 −+−+−+ = ( ) ( )ii 33252 +−++ = ( ) ( )ii 33247.4 +−++ 
 
= ( ) ( )ii 32347.4 ++− = i547.1 + 
Mastering Algebra - Advanced Level 2.6 Math Operations Involving Complex Numbers 
Hamilton Education Guides 160 
Example 2.6-9 
( ) ( )ii 3185 −−− = ( ) ( )ii 3185 +−+− = ( ) ( )ii 32215 +−+− = ( )i73.183.24 +−+ = i1.14 − 
Example 2.6-10 
( )[ ] ( )iii 3513272 −−++ = ( )[ ] ( )iii 3513332 +−+++ = ( ) ( )ii 351342 +−++ 
 
= ( ) ( )ii 353412 ++− = i391+ = i59.151+ 
 
Case II Multiplication of Complex Numbers 
In general, two complex numbers are multiplied by one another in the following way: 
( ) ( )idciba +⋅+ = 2ibdibciadac +++ = bdibciadac −++ = ( ) ( )ibcadbdac ++− 
Note that the distributive property of multiplication is also valid for complex numbers. In 
addition, note that multiplication of two complex numbers is similar to the multiplication of two 
binomials using the Foil Method (see Section 1.2b, Case II). 
Example 2.6-11 
( )( )ii 2354 −+ = ( ) ( ) ( ) ( )iiii 25352434 −×+×+−×+× = 21015812 iii −+− = ( )11015812 −×−++− ii 
 
= 1015812 ++− ii = ( ) ( )i1581012 +−++ = i722 + 
Example 2.6-12 
( )( )ii +−−− 532 = ( ) ( ) ( ) ( )iiii ×−+−×−+×−+−×− 353252 = 2315210 iii −+− = ( )1315210 −×−++− ii 
 
= 315210 ++− ii = ( ) ( )ii 152310 +−++ = i1313 + 
Example 2.6-13 
( )( )ii 3074 +− = ( ) ii 374 ×− = iii 3734 ×−× = 22112 ii − = 2112 +i = i1221+ 
Example 2.6-14 
( )( )ii −− 235 = ( ) ( ) ( ) ( )iii −×−+−×+−×+× 3321525 = 236510 iii +−− = 36510 −−− ii = ( ) ( )i65310 −−+− 
 
= i117 − 
Example 2.6-15 
( )( )ii 5252 −+ = ( ) ( ) ( ) ( )iiii 55525222 −×+−×+−×+× = 22510104 iii −+− = ( ) 12510104 −×−+−+ i 
 
= 2504 ++ = 29 
Example 2.6-16 
( )( )ii 2553 −+ = ( ) ( ) ( ) ( )iiii 25552353 −×+×+−×+× = 210552315 iii −+− 
 
= ( ) ( )i55231015 +−++ = ( ) ( )i18.1124.416.315 +−++ = i94.616.18 + 
Mastering Algebra - Advanced Level 2.6 Math Operations Involving Complex Numbers 
Hamilton Education Guides 161 
Example 2.6-17 
( )[ ]( )iii −+− 255 3 = ( )[ ]( )iii −+−− 255 = ( )( )iii −− 2525 2 = ( )( )ii −+ 2255 = ( ) ( ) ( )225525 ×+−×+× ii 
 
( )ii −×+ 25 = 22550510 iii −+− = ( ) ( )i5052510 +−++ = i4535 + 
Example 2.6-18 
( )( )ii 2284 +− = ( )( )ii 22224 +− = ( ) ( ) ( ) ( )iiii 2222222424 ×−+×−+×+× = i248+ 
 
2424 ii −− = 248 i− = 48+ = 12 
Example 2.6-19 
( )( )ii −−− 2236 = ( ) ( ) ( ) ( )iiii −×−+−×−+−×+−× 32236226 = 23266122 iii ++−− 
 
= 326634 −+−− ii = ( ) ( )i266334 +−+−− = ( ) ( )i49.845.2393.6 +−+−− = i04.693.9 +− 
Example 2.6-20 
( )( )[ ]iii 53522 −− = ( )( )ii 5352 −−− = ( )( )ii 5352 −+− = ( ) ( ) ( ) ( )iiii 55355232 −×+×+−×−+×− 
 
= 22515106 iii −++− = 2515106 +++− ii = ( ) ( )i1510256 +++− = i2519 + 
 
Case III Division of Complex Numbers 
In general, two complex numbers are divided by one another in the following way: 
idc
iba
+
+ = 
idc
iba
idc
iba
−
−
×
+
+ = ( )( )( )( )idcidc
idciba
−+
−+ = 
222
2
idicdicdc
ibdibciadac
−−+
−+− = 
22 dc
bdibciadac
+
++− = ( ) ( )
22 dc
iadbcbdac
+
−++ 
= ( )
2222 dc
ibcad
dc
bdac
+
+
+
+
+ 
Note that division of two complex numbers that contain i in the denominator is similar to 
rationalizing the denominator for radical expressions where the numerator and the denominator 
are multiplied by the conjugate of the denominator (see Section 1.2b, Case V). 
Example 2.6-21 
i
i32 + = 
i
i
+
+
0
32 = 
i
i
i
i
−
−
×
+
+
0
0
0
32 = 
i
i
i
i
−
−
×
+ 32 = ( )
ii
ii
×−
−×+ 32 = 2
232
i
ii
−
−− = ( )( )1
132
−−
−×−+− i = i23 − 
Example 2.6-22 
i
i
−
+−2 = 
i
i
i
i
+
+
×
−
+−
0
0
0
2 = 
i
i
i
i
×
−
+−2 = ( )
ii
ii
×−
×+−2 = 2
22
i
ii
−
+− = ( )1
12
−−
−− i = 
1
21 i−− = i21 −− 
Example 2.6-23 
i
i
23
52
+
− = 
i
i
i
i
23
23
23
52
−
−
×
+
− = 2
2
4669
101546
iii
iii
−+−
+−− = 
49
10196
+
−− i = 
13
194 i−− = i
13
19
13
4
−− = i46.131.0 −− 
Mastering Algebra - Advanced Level 2.6 Math Operations Involving Complex Numbers 
Hamilton Education Guides 162 
Example 2.6-24 
i
i
41+
 = 
i
i
i
i
41
41
41 −
−
×
+
 = 2
2
4441
4
iii
ii
−+−
− = 241
4
i
i
−
+ = 
41
4
+
+ i = 
5
4 i+ = i
5
1
5
4
+ = i2.08.0 + 
Example 2.6-25 
i
i
32
4
+
+ = 
i
i
i
i
32
32
32
4
−
−
×
+
+ = 2
2
9664
32128
iii
iii
−+−
−+− = ( )
94
1038
+
−+ i = 
13
1011 i− = i
13
10
13
11
− = i77.085.0 − 
Example 2.6-26 
i
i
52
2
+
+ = 
i
i
i
i
52
52
52
2
−
−
×
+
+ = 2
2
2525252
521022
iii
iii
−+−
−+− = 
252
521022
+
++− ii = 
( ) ( )
27
210522 i+−++
 
 
= ( ) ( )
27
41.110583.2 i+−++ = 
27
59.883.7 i− = i
27
59.8
27
83.7
− = i32.029.0 − 
Example 2.6-27 
i
i
253
23
−
+− = 
i
i
53
23
−
+− = 
i
i
i
i
53
53
53
23
+
+
×
−
+− = 2
2
2535353
2561533
iii
iii
−−+
++−− = 
253
2561533
+
−+−− ii 
 
= ( ) ( )
28
6152533 i+−+−− = ( ) ( )
28
45.21507.72.5 i+−+−− = 
28
55.1227.12 i−− = i45.044.0 −− 
Example 2.6-28 
383
51
i
i
−
−
 = 
i
i
83
51
+
−
 = 
i
i
i
i
83
83
83
51
−
−
×
+
−
 = 2
2
6424249
585383
iii
iii
−+−
+−−
 = 
649
585383
+
−−− ii
 = 
( ) ( )
73
538583 i+−−
 
 
= ( ) ( )
73
71.6889.173 i+−− = 
73
71.1489.14 i−− = i
73
71.14
73
89.14
−− = i2.02.0 −− 
Example 2.6-29 
5
53
i
i− = 
iii
i
⋅⋅
−
22
53 = 
i
i
×−×−
−
11
53 = 
i
i53− = 
i
i
+
−
0
53 = 
i
i
i
i
−
−
×
+
−
0
0
0
53 = 
i
i
i
i
−
−
×
−53 = 2
253
i
ii
−
+− = ( )1
53
−−
−− i 
 
= 
1
35 i−− = i35−− 
Example 2.6-30 
i
i
32
4
+
+ = 
i
i
i
i
32
32
32
2
−
−
×
+
+ = 2
2
932324
32324
iii
iii
−+−
−+− = 234
32324
i
ii
−
++− = 
( ) ( )
34
23234
+
+−++ i
 
 
= ( ) ( )
7
23234 i+−++ = ( ) ( )
7
246.373.14 i+−++ = 
7
46.173.5 i− = i
7
46.1
7
73.5
− = i21.082.0− 
Mastering Algebra - Advanced Level 2.6 Math Operations Involving Complex Numbers 
Hamilton Education Guides 163 
Case IV Mixed Operations Involving Complex Numbers 
In this case combining addition, subtraction, multiplication, and division of complex numbers are 
addressed. Note that math properties discussed in Cases I through III are also applicable here. 
The following show examples of mixed operations involving complex numbers. 
Example 2.6-31 
i
i
i
i
+
+
×
+
+
3
2
1
32 = ( ) ( )( ) ( )ii
ii
+×+
+×+
31
232 = 2
2
33
3624
iii
iii
+++
+++ = 
133
3624
−++
−++
ii
ii = ( ) ( )( ) ( )ii
ii
313
6234
++−
++−
= 
( ) ( )
( ) ( )i
i
3113
6234
++−
++− 
 
= 
i
i
42
81
+
+ = ( ) ( )( ) ( )ii
ii
4242
4281
−×+
−×+ = 2
2
16884
321642
iii
iii
−+−
−+− = 
164
321642
+
++− ii = 
20
1234 i+ = i
20
12
20
34
+ = i6.07.1 + 
Example 2.6-32 
i
i
i
i 3221 −
+
− = ( ) ( )
i
ii 3221 −+− = ( ) ( )
i
ii 3221 −−++ = 
i
i53− = 
i
i
+
−
0
53 = 
i
i
i
i
−
−
×
+
−
0
0
0
53 = ( )
ii
ii
−×
−×−53 
 
= 2
253
i
ii
−
+− = 
1
53 −− i = i35−− 
Example 2.6-33 
i
i
i
i
−
+
÷
+
+
1
1
31
2 = 
i
i
i
i
+
−
×
+
+
1
1
31
2 = ( ) ( )( ) ( )ii
ii
+×+
−×+
131
12 = 2
2
331
22
iii
iii
+++
−+− = 
331
122
−++
++−
ii
ii = ( ) ( )( ) ( )i
i
3131
1212
++−
+−++ = 
i
i
42
3
+−
− 
 
= 
i
i
i
i
42
42
42
3
−−
−−
×
+−
− = ( )( )( )( )ii
ii
4242
423
−−+−
−−− = 2
2
16884
42126
iii
iii
−−+
++−− = 
164
42126
+
−+−− ii = ( ) ( )
20
21246 i+−+−− 
 
= 
20
1010 i−− = i
20
10
20
10
−− = i5.05.0 −− 
Example 2.6-34 
( )( )[ ] ( )iii 21132 −+−+ = ( ) ( )iiii 213322 2 −+−+− = ( ) ( )iii 213322 −+++− = ( ) ( )[ ] ( )ii 213232 −++−++ 
 
= ( ) ( )ii 215 −++ = ( ) ( )ii 215 −++ = i−6 
Example 2.6-35 
( ) ( )iiii 211 32 +−+ = ( ) ( )iii 211 +++− = ( ) ( )221 iii ++−− = ( ) ( )21 −+−− ii = ( ) ( )ii +−+−− 21 = 3− 
Example 2.6-36 
( ) ( )[ ] 5225 iii ×−−− = ( ) ( )[ ] ( )iiiii ⋅⋅×+−+− 22225 = ( ) ( )[ ] iii ×+−+− 225 = ( ) ii ×−3 = 23 ii − = i31+ 
Example 2.6-37 
ii
i
21
1
1
2
+
÷
+
− = 
1
21
1
2 i
i
i +
×
+
− = ( ) ( )( ) 11
212
×+
+×−
i
ii = 
i
iii
+
−−+
1
242 2 = 
i
ii
+
+−+
1
242 = 
( ) ( )
i
i
+
−++
1
1422 = 
i
i
+
+
1
34 
 
Mastering Algebra - Advanced Level 2.6 Math Operations Involving Complex Numbers 
Hamilton Education Guides 164 
= 
i
i
i
i
−
−
×
+
+
1
1
1
34 = 
( )( )
( )( )ii
ii
−+
−+
11
134
 = 2
2
1
3344
iii
iii
−+−
−+− = 
11
3344
+
++− ii = ( ) ( )
2
3434 i+−++ = 
2
7 i− = i5.05.3 − 
Example 2.6-38 
i
i
i
i +
×
−
2
1
4
 = 
i
i
i
ii +
×
−
× 2
1
22
 = 
i
i
i
+
×
−
−×− 2
1
11 = 
i
i
i
+
×
−
2
1
1 = ( )( ) ii
i
×−
+×
1
21 = 2
2
ii
i
−
+ = 
i
i
+
+
1
2 = 
i
i
i
i
−
−
×
+
+
1
1
1
2 
 
= ( ) ( )( ) ( )ii
ii
−×+
−×+
11
12 = 2
2
1
22
iii
iii
−+−
−+− = 
11
122
+
++− ii = ( ) ( )
2
1212 i+−++ = 
2
3 i− = i
2
1
2
3
− = i5.05.1 − 
Example 2.6-39 
( )( )
i
ii
21
243
−
+− = 
i
iii
21
4836 2
−
−−+ = 
i
ii
21
4836
−
+−+ = ( ) ( )
i
ii
21
8346
−
−++ = 
i
i
21
510
−
− = 
i
i
i
i
21
21
21
510
+
+
×
−
− 
 
= ( ) ( )( ) ( )ii
ii
2121
21510
+×−
+×− = 2
2
4221
1052010
iii
iii
−−+
−−+ = 
41
1052010
+
+−+ ii = ( ) ( )
5
5201010 ii −++ = 
5
1520 i+ = i34 + 
Example 2.6-40 
2
3
4
i
i
i
÷
−
 = 
14
2i
i
i
÷
−
− = 2
1
4 ii
i
×
−
− = 
1
1
4 −
×
−
−
i
i = ( ) 14
1
−×−
×−
i
i = ( )i
i
−−
−
4
 = 
i
i
−4
 = 
i
i
i
i
+
+
×
− 4
4
4
 
 
= ( )( )( )ii
ii
+−
+
44
4 = 2
2
4416
4
iii
ii
−−+
+ = 216
14
i
i
−
− = 
116
41
+
+− i = 
17
41 i+− = i
17
4
17
1
+− = i24.006.0 +− 
Example 2.6-41 
( )
i
ii
3
231 +×+ = ( )
i
ii
3
2
1
31 +
×
+ = ( ) ( )
i
ii
31
231
×
+×+ = 
i
iii
3
362 2+++ = 
i
ii
3
362 −++ = 
( ) ( )
i
i
3
6132 ++−
 = 
i
i
30
71
+
+− 
 
= 
i
i
i
i
30
30
30
71
−
−
×
+
+− = ( )
ii
ii
33
371
−×
−×+− = 2
2
9
213
i
ii
−
− = 
9
213 +i = i
9
3
9
21
+ = i33.033.2 + 
Example 2.6-42 
( )( )
( )( )ii
ii
+−
+−
12
234 = 2
2
22
3648
iii
iii
−−+
−−+ = 
122
3648
+−+
+−+
ii
ii = ( ) ( )( ) ( )ii
ii
−++
−++
212
6438 = 
i
i
+
−
3
211 = 
i
i
i
i
−
−
×
+
−
3
3
3
211 
 
= ( ) ( )( ) ( )ii
ii
−×+
−×−
33
3211 = 2
2
339
261133
iii
iii
−+−
+−− = 
19
261133
+
−−− ii = ( ) ( )
10
611233 ii −−+− = 
10
1731 i− = i7.11.3 − 
Example 2.6-43 
ii 32
1
1
3
−
×
+
 = ( )( )ii 321
13
−+
× = 23232
3
iii −+−
 = 
3232
3
++− ii
 = ( ) ( )i2332
3
+−++
 = 
i−5
3 = 
i
i
i +
+
×
− 5
5
5
3 
 
( )
( )( )ii
i
+−
+
55
53 = 25525
315
iii
i
−−+
+ = 225
315
i
i
−
+ = 
125
315
+
+ i = 
26
315 i+ = i
26
3
26
15
+ = i12.058.0 + 
Mastering Algebra - Advanced Level 2.6 Math Operations Involving Complex Numbers 
Hamilton Education Guides 165 
Example 2.6-44 
biabia −
+
−
52 = 
bia −
+ 52 = 
bia
bia
bia +
+
×
−
7 = ( )( )( )biabia
bia
+−
+7 = ( ) 222
7
ibabiabia
bia
−−+
+ = ( )222
7
iba
bia
−
+ 
 
= ( )22
7
ba
bia
+
+ = i
ba
b
ba
a
2222
77
+
+
+
 
Example 2.6-45 
ii −
−
+ 1
4
32
1 = ( )[ ] ( )[ ]( )( )ii
ii
−+
+×−−×
132
32411 = ( ) ( )23322
1281
iii
ii
−+−
−−+− = ( ) ( )
3322
1281
++−
−−+−
ii
ii = ( ) ( )ii
i
3232
137
+−++
−− 
 
= 
i
i
+
−−
5
137 = 
i
i
i
i
−
−
×
+
−−
5
5
5
137 = ( )( )( )( )ii
ii
−+
−−−
55
5137 = 2
2
5525
1365735
iii
iii
−+−
+−+− = 
125
1365735
+
−−+− ii 
 
= ( ) ( )
26
6571335 ii −+−− = 
26
5848 i−− = i
26
58
26
48
−− = i23.285.1 −− 
Example 2.6-46 
i
i
i 23
32
1
4
−
+
+
+
 = ( )[ ] ( ) ( )[ ]( )( )ii
iii
231
132234
−+
+×++−× = ( )[ ] ( ) ( )[ ]( )( )ii
iii
231
132234
−+
+×++−× = ( ) ( )2
2
2323
3322812
iii
iiii
−+−
++++− 
 
= ( ) ( )( ) ( )ii
iii
3223
3283212
+−++
++−+−+ = 
i
i
+
−
5
311 = 
i
i
i
i
−
−
×
+
−
5
5
5
311 = ( )( )( )( )ii
ii
−+
−−
55
5311 = 2
2
5525
3151155
iii
iii
−+−
+−− 
 
= 
125
3151155
+
−−− ii = ( ) ( )
26
1511355 ii −−+− = 
26
2652 i− = i
26
26
26
52
− = i−2 
Example 2.6-47 
ii
i 1
2
13 +




 ÷ = 
i
ii 1
1
2
1
3
+







× = 
i
ii 1
11
23
+
×
× = 
i
i 1
1
2 4
+ = 
i
1
1
112
+
−×−× = 
i
1
1
2
+ = ( ) ( )
i
i
×
×+×
1
112 = 
i
i 12 + 
 
= 
i
i
+
+
0
12 = 
i
i
i
i
−
−
×
+
+
0
0
0
12 = ( )
ii
ii
−×
−×+12 = 2
22
i
ii
−
−− = 
1
2 i− = i−2 
Example 2.6-48 
3
2
3
1
1 ii
i
−
×
−
 = 
ii
i
+
×
− 3
1
1
2
 = ( )( )ii
i
+−
×
31
12 = 233
1
iii −−+
− = 
133
1
+−+
−
ii
 = ( ) ( )i3113
1
−++
− = 
i24
1
−
− 
 
= 
i
i
i 24
24
24
1
+
+
×
−
− = ( )( )( )ii
i
2424
241
+−
+×− = 248816
24
iii
i
−−+
−− = 
416
24
+
−− i = 
20
24 i−− = i
20
2
20
4
−− = i1.02.0 −− 
Example 2.6-49 
ii
i
32
11
+
÷
+ = 
1
321 i
i
i +
×
+ = ( ) ( )
1
321
×
+×+
i
ii = 
i
iii 23232 +++ = 
i
ii 3232 −++ = ( ) ( )
i
i2332 ++− 
 
Mastering Algebra - Advanced Level 2.6 Math Operations Involving Complex Numbers 
Hamilton Education Guides 166 
= 
i
i
+
+−
0
51 = 
i
i
i
i
−
−
×
+
+−
0
0
0
51 = ( )
ii
ii
−×
−×+− 51 = 2
25
i
ii
−
− = ( )1
5
−−
+i = 
1
5+i = i+5 
Example 2.6-50 
( )( )[ ]
2
1254
i
ii ++− = ( )
1
151048 2
−
+−−+ iii = ( ) 151048 −+−+ ii = ( ) ( )ii 104158 −+−+ = i612 − 
 
Practice Problems – Math Operations Involving Complex Numbers 
Section 2.6 Case I Practice Problems – Add or subtract the following complex numbers: 
a. ( ) ( )ii 5824 −++ = 
 
b. ( ) ( )ii 4537 +−−− = c. ( ) ( ) ( )[ ]iii −−−++ 65274 = 
d. ( ) ( )254352 −−−−+ = e. ( )[ ] ( )iii 531552 3 +−++ = f. ( ) ( )ii 2163 +−− = 
 
Section 2.6 Case II Practice Problems – Multiply the following complex numbers by one another: 
a. ( )( )ii 6325 −+ = 
 
b. ( )( )ii +−−− 726 = c. ( )[ ]( )iii +− 353 6 = 
d. ( )( )ii 3593 +− = e. ( )( )ii +−− 535 = f. ( )( )[ ]iii 24423 −+ = 
 
Section 2.6 Case III Practice Problems– Divide the following complex numbers by one another: 
a. 
i
i
35
41
+
− = b. 
i
i
81
3
−
 = c. 
i
i
43
2
+
− = 
d. 
i
i
365
22
−
+− = e. 
581
33
i
i
−
− = f. 71
23
i
i
−
+ = 
 
Section 2.6 Case IV Practice Problems – Simplify the following expressions involving complex 
numbers: 
a. ( ) ( )iiii 5221 43 −+− = b. 
ii
i
31
1
1
54
+
÷
−
+ = c. 
ii
i
+
×
+
−
1
1
32
1 = 
 
d. 
i
ii
21
428
+
−
÷ = e. ( )( )( )( )ii
ii
3241
152
++
−+ = f. ( )
i
i
−
−+
1
185 = 
 
g. ( )
i
i
31
224
−
÷+ = h. ( )
i
ii
+
−
+−
1
3152 = i. 
i
i
i
i
+
−
÷
+
1
4232 = 
 
 
Mastering Algebra - Advanced Level Quick Reference to Chapter 3 Problems 
Hamilton Education Guides 167 
Chapter 3 
Matrices 
 
Quick Reference to Chapter 3 Problems 
3.1 Introduction to Matrices ........................................................................................... 169 






−
=× 131
321
32A = ; 










−=×
010
111
301
33B = ; 










−
−
−
=×
6402
3110
5321
43C = 
3.2 Matrix Operations ..................................................................................................... 173 
Case I - Matrix Addition and Subtraction, p. 173 
1 2
3 5
2 4
3 1





 + −





 = ; 










−
−+









 −
351
032
971
832
651
1310
 = ; 
1 3 6
1 0 2
5 2 4
3 5 2
0 4 8
5 3 6
−
−










−
−
−
−










 = 
Case II - Matrix Multiplication, p. 178 
−








 −










1 0 1
3 1 0
1 0 2
0 1 2
3 0 1
1 1 0
 = ; 
1 0 2
3 1 4
1 2
1 3
2 5
−





 −










 = ; 
1 0 1
2 3 4
5 1 1
1
2
5
−
−
−




















=










x
y
z
 = 
3.3 Determinants .............................................................................................................. 185 
( )
102
001
300
−
=Aδ = ; ( )
52
31
4
−
×=Bδ = ; ( )
142
031
300
−
=Cδ = 
3.4 Inverse Matrices ......................................................................................................... 198 
1
13
11
21
23
−













 −





 = ; 
1 0
2 3
1 3
0 1
1 1





 ⋅ −






− −
 = ; 
1
32
01
10
31
−














⋅





−
 = 
3.5 Solving Linear Systems.............................................................................................. 210 
Case I - Solving Linear Systems Using the Addition Method, p. 210 
15
132
=−
−=+−
yx
yx
 = ; 
122
122
2
=++
−=−
−=+
zyx
zx
yx
 = ; 
23
03
12
=−+
=++−
=−+
zyx
zyx
zyx
 = 
Case II - Solving Linear Systems Using the Substitution Method, p. 214 
02
132
=−
=+−
yx
yx
 = ; 
04
123
1
=+
=−
−=+
zy
zx
yx
 = ; 
122
4
332
−=++
−=+
=+−
zyx
zx
zyx
 = 
 
Mastering Algebra - Advanced Level Quick Reference to Chapter 3 Problems 
Hamilton Education Guides 168 
Case III - Solving Linear Systems Using the Inverse Matrices Method, p. 218 
x y z
x y z
x y z
+ − = −
− + =
+ − = −
3 3 1
2 3 2 3
2 2
 = ; 
x y
x z
y z
+ = −
− = −
+ =
3
2 4
2 2 1
 = ; 
x y z
x y z
x y z
+ − = −
− + + =
+ − =
3 1
2 3 0
2 2
 = 
Case IV - Solving Linear Systems Using Cramer’s Rule, p. 224 
− + =
− =
x y
x y
2 1
4 0
 = ; 
x y
x z
y z
+ =
− =
+ =
0
2 1
2 0
 = ; 
1622
3
23
−=+−
−=−
=+−
zyx
zx
zyx
 = 
Case V - Solving Linear Systems Using the Gaussian Elimination Method, p. 228 
2 3 6
4 2
x y
x y
+ =
− = −
 = ; 
2 3 1
3 2 0
2 1
x y z
x z
x y
− + = −
+ =
− =
 = ; 
2 3 1
3 0
2 2 3 2
x z
x y
x y z
+ = −
+ =
− + = −
 = 
Case VI - Solving Linear Systems Using the Gauss-Jordan Elimination Method, p. 234 
0
3
23
22
=−
=−
yx
yx
 = ; 
3 2 1
0
2 3 2
x z
x y z
x y
− = −
− + =
+ = −
 = ; 
0
13
0
=+
−=−
=−
yx
yx
zx
 = 
 
 
Hamilton Education Guides 169 
Chapter 3 - Matrices 
 
The objective of this chapter is to improve the student’s ability to solve problems involving 
matrices. Matrices are introduced in Section 3.1. How to add, subtract, and multiply matrices 
are introduced in Section 3.2. Calculating minors and cofactors and their use in finding the 
determinant of a matrix are addressed in Section 3.3. Computing the inverse of a square matrix 
and the steps in finding an inverse matrix are addressed in Section 3.4. In Section 3.5, solving 
linear systems using various methods such as the Addition, Substitution, Inverse Matrix, 
Cramer’s Rule, Gaussian Elimination, and Gauss-Jordan Elimination methods are discussed in 
Section 3.6. Each section is concluded by solving examples with practice problems to further 
enhance the student’s ability. Students are encouraged to gain a thorough knowledge of the 
matrix operations and learn how to find the determinant and inverse of a matrix. An in depth 
knowledge of matrix properties will greatly simplify solutions to linear systems of equations and 
introduce students to various methods used in solving these systems. 
3.1 Introduction to Matrices 
A matrix is defined as a rectangular array of numbers which are called the entries or elements of 
the matrix. In general, a matrix A with m rows and n columns is represented as 
A
a a a a
a a a a
a a a a
a a a a
n
n
n
m m m mn
=
















11 12 13 1
21 22 23 2
31 32 33 3
1 2 3



    

 
1
2
3
st
nd
rd
th
row
row
row
m row
 
 
..
3
.
2
.
1
col
n
colcolcol
thrdndst
 
Note that each matrix entry is read first by its location relative to the row and second by its 
location relative to the column. For example, a11 is read as the entry in the first row and the first 
column, a32 is read as the entry in the third row and the second column, am3 is read as the entry 
in the mth row and the third column, amn is read as the entry in the m
th row and the nth column, 
etc. The order or dimension of a matrix is given as m n× (reads as “ m by n ”). For example, 
1 0
2 1





 , 
1 2 3
1 3 1−





 , 
1 0 3
1 1 1
0 1 0
−










, 
1
2
3−










, 
1 2 3 5
0 1 1 3
2 0 4 6
−
−
−










, 
1 0
3 6
1 4
2 1
−












, and 
1 1 3 5
0 3 2 5
1 1 5 0
2 3 4 1
−
−
−












 
are a 2 2× (reads as a “ 2 by 2 ”), a 2 3× (reads as a “ 2 by 3 ”), a 3 3× (reads as a “ 3 by 3 ”), a 
3 1× (reads as a “ 3 by 1”), a 3 4× (reads as a “ 3 by 4 ”), a 4 2× (reads as a “ 4 by 2 ”), and a 
4 4× (reads as a “ 4 by 4 ”) matrices, respectively. In addition, a matrix consisting of a single 
row is referred to as a row matrix and a matrix consisting of a single column is referred to as a 
column matrix. Finally, note that in general matrices are represented by capital letters. For 
example, 
Mastering Algebra - Advanced Level 3.1 Introduction to Matrices 
Hamilton Education Guides 170 
matrices 1 0
3 5−





 and 
3
5





 are shown as A or A2 2× and B or B2 1× matrices. 
In the following sections, we will learn about various matrix operations, determinants, inverse 
matrices, as well as different methods for solving linear systems of equations. However, we first 
need to learn about several types of matrices known as: equal, transpose, zero, square, diagonal, 
identity, coefficient, and augmented matrices. 
Definition 3.1-1: Equal Matrices 
Two matrices are said to be equal if and only if 1) they both are of the same order and 2) their 
corresponding entries are equal. 
Example 3.1-1: The following matrices are equal to each other. 
a. 1 5
2 3





 = 
1 15
3
4 3








 b. 2 0
1 9−





 = 
8
4
02
2
3−










 c. 
3
4
3
0 6 1
−
−







.
 = 
0 75 9
3
3
5
2
2
. −
−










 d. 3
5





 = 
6
2
25








 
Example 3.1-2: The following matrices are not equal to each other. 
a. 3 5
0 1





 ≠ 
3 0
5 1





 b. 
−





3 2
0 5
 ≠ −





3 2
5 0
 c. 3 1
2 4





 ≠ 
9 5
5
4
2
8
4










 d. 0
5





 ≠ [ ]0 5 
Example 3.1-3: Given the following equal matrices solve for the unknowns. 
a. x
y
−
−






2
3 5
 = 3
2 5
z
−





 b. 
2 2 2
3 3 2
−
− +






a b
c d
 = e b
f
6 5
3 3 4
− +
+





 
Solution: 
a. Equating each corresponding element we obtain: 
x = 3 − =2 z or z = −2 y − = −3 2 or y = 1 
b. Equating each corresponding element we obtain: 
2 = e or e = 2 − =2 6a or a = −3 2 5b b= − + or 3 5b = ; b = 5
3
 
c − =3 3 or c = 6 3 3= + f or f = 0 d + =2 4 or d = 2 
Definition 3.1-2: Transpose of a Matrix 
The transpose of a matrix is a matrix in which the rows and columns are interchanged. The 
transpose of a matrix is denoted by At . 
Example 3.1-4: Find the transpose of the following matrices. 
a. 1 5
2 3






t
 = 1 2
5 3





 b. 
1 5 6
2 0 3






t
 = 
1 2
5 0
6 3










 c. 1 0
0 1






t
 = 1 0
0 1





 d. 
3
5
2










t
 = [ ]3 5 2 
Definition 3.1-3: Zero Matrix 
A matrix with zero entries is referred to as a zero matrix. A zero matrix is generally denoted by 
the zero symbol 0 or a zero with subscript indicating the number of rows ( )m and columns ( )n , 
i.e., nm×0 . For example, 2 2× , 3 1× , and 2 3× zero matrices are represented as: 
Mastering Algebra - Advanced Level 3.1 Introduction to Matrices 
Hamilton Education Guides 171 
02 2× = 
0 0
0 0





 03 1× = 
0
0
0










 02 3× = 
0 0 0
0 0 0





 
Definition 3.1-4: Square Matrix 
A matrix having the same number of rows as columns is referred to as a square matrix. For 
example, 1 2
4 0
−
−





 , 
−
−










1 0 5
3 4 2
0 2 7
 , and 
−
−












3 4 7 2
1 0 3 5
2 5 7 10
1 0 1 4
 are 2 2× , 3 3× , and 4 4× matrices. 
Definition 3.1-5: Diagonal Matrix 
A diagonal matrix is a square matrix in which only the diagonal matrix entries are not equal to 
zero. For example, A2 2× = 
1 0
0 5





 and B3 3× = 
−









1 0 0
0 4 0
0 0 7
 are diagonal matrices. 
Definition 3.1-6: Identity Matrix 
An identity matrix is a matrix in which only the diagonal matrix entries are equal to one. An 
identity matrix is generally denoted by the symbol I . I2 2× = 
1 0
0 1





 , I3 3× = 
1 0 0
0 1 0
0 0 1










 are 
examples of the identity matrix. 
Definition 3.1-7: Coefficient Matrix 
A coefficient matrix is a matrix in which its first, second, third, fourth, etc. columns are formed 
from the coefficients of unknown variables x , y , z , w , etc. presented as a system of linear 
equations. For example, given the system of linear equations 
1023
352
=−
=+
yx
yx the matrix 





− 23
52 is 
called the coefficient matrix. 
Definition 3.1-8: Augmented Matrix 
An augmented matrix is a coefficient matrix which includes the columns consisting the right 
hand sides of the linear equations separated by a dashed line. For example, given the system of 
linear equations 
1023
352
=−
=+
yx
yx the matrix 





− 1023
352

 is called the augmented matrix. 
Having identified different types of matrices, in the following section we will address various 
matrix operations which includes the addition, subtraction, and multiplication of matrices. 
Section 3.1 Practice Problems - Introduction to Matrices 
1. State the order and find the transpose of each matrix. 
a. 1 0
2 3





 b. 
1 4
0 2





 c. 
1 2 3
1 0 1
−
−





 d. 
1 0
2 1
1 3− −










 
Mastering Algebra - Advanced Level 3.1 Introduction to Matrices 
Hamilton Education Guides 172 
e. 
2 3 5
1 4 1
3 0 2
−










 f. 
1
2
3










 g. [ ]1 1 2− h. 
1 0 1 2
2 3 1 0
1 0 2 3
−
− −










 
2. Given A =
−
−










1 2 3
0 1 1
2 3 0
, find a12 , a23 , a33 , a22 , a31 , a21 , and a11 . 
3. Given B =
−
− −










1 1 2 3
0 1 1 3
3 1 2 5
, find b12 , b21 , b32 , b33 , b23 , b24 , b34 , and b14 . 
4. Given A x2 3 , B x3 1 , A x1 2 , A x3 3 , B x2 4 , B x1 4 , B x4 1 , and A x3 4 write a matrix that corresponds to 
 the order given. 
5. State if the given paired matrices are equal to each other. 
a. 1 0
1 3
4 0
2
2
6
2
−





 = −








?
 b. −




 =
−







1 2
3 6
2
2
4
8
6 12
?
 c. −




 =
−











1 0 2
1 2 3
1 0 16
8
25
5
6
3
9
?
 
d. 
3
4
4
0 8 1
0 75 8
2
8
10
4
4
−
−








=
−








.
? .
 e. 1 0
2 3
2
2
0
2
4 9
3
−





 =
−










?
 f. 
1 4
3 5
1 2
1 16
27
9
25
5
6
3
2
−










=
−
















?
 
Mastering Algebra - Advanced Level 3.2 Matrix Operations 
Hamilton Education Guides 173 
3.2 Matrix Operations 
In the previous section we defined different types of matrices. In this section, we will learn about 
matrix addition and subtraction (Case I) and discuss how matrices are multiplied by one another 
(Case II). 
Case I Matrix Addition and Subtraction 
To add matrices, we simply add or subtract the corresponding entries. Note that to add matrices 
the matrices must have the same order, i.e., we can only add a 2 2× or a 3 3× matrix with another 
2 2× or 3 3× matrix. Hence, the sum of two matrices of different orders is not defined. 
In general, matrices are added or subtracted in the following way: 
a a a
a a a
a a a
n
n
m m mn
11 12 1
21 22 2
1 2


  













 + 
b b b
b b b
b b b
n
n
m m mn
11 12 1
21 22 2
1 2


  













 = 












+++
+++
+++
mnmnmmmm
nn
nn
bababa
bababa
bababa




2211
2222222121
1112121111
 = 
c c c
c c c
c c c
n
n
m m mn
11 12 1
21 22 2
1 2


  













 
 












mnmm
n
n
aaa
aaa
aaa




21
22221
11211
 - 












mnmm
n
n
bbb
bbb
bbb




21
22221
11211
 = 












mnmm
n
n
aaa
aaa
aaa




21
22221
11211
 + 












−−−
−−−
−−−
mnmm
n
n
bbb
bbb
bbb




21
22221
11211
 
 
= 












−−−
−−−
−−−
mnmnmmmm
nn
nn
bababa
bababa
bababa




2211
2222222121
1112121111
 = 
d d d
d d d
d d d
n
n
m m mn
11 12 1
21 22 2
1 2


  













 
Example 3.2-1: Add or subtract the following square matrices. 
a. 1 2
3 5
2 2
3 4





 +
−




 = b. 
1 2 5
3 1 4
1 3 3
1 0 5−





 +
−
−





 = 
c. 3 3 2
1 0 5
2 5 1
3 2 0
− −




 − − −





 = d. 
1 3 6
1 0 2
5 2 4
3 5 2
0 4 8
5 3 6
−
−










−
−
−
−










 = 
Solutions: 
a. 
1 2
3 5
2 2
3 4





 +
−




 = 
1 2 2 2
3 3 5 4
+ −
+ +





 = 
3 0
6 9





 
 
b. 
1 2 5
3 1 4
1 3 3
1 0 5−





 +
−−





 = 
1 1 2 3 5 3
3 1 1 0 4 5
+ + −
− + − +





 = 
2 5 2
2 1 1





 
 
c. 
3 3 2
1 0 5
2 5 1
3 2 0
− −




 − − −





 = 
3 3 2
1 0 5
2 5 1
3 2 0
− −




 +
− − −




 = 
3 2 3 5 2 1
1 3 0 2 5 0
− − − − −
+ + +





 = 
1 8 3
4 2 5
− −





Mastering Algebra - Advanced Level 3.2 Matrix Operations 
Hamilton Education Guides 174 
d. 
1 3 6
1 0 2
5 2 4
3 5 2
0 4 8
5 3 6
−
−










−
−
−
−










 = 
1 3 6
1 0 2
5 2 4
3 5 2
0 4 8
5 3 6
−
−










+
− −
−
− −










 = 
1 3 3 5 6 2
1 0 0 4 2 8
5 5 2 3 4 6
+ − −
− + + −
− − + −










 = 
4 2 4
1 4 6
0 1 2
−
− −
−










 
Properties of Matrix Addition 
a. Matrix addition is both associative and commutative. This implies that matrices of the same 
dimension can be added in any order. For example, if A , B , and C are m n× matrices, then: 
 ( ) ( )A B C A B C+ + = + + Associative property of addition 
 A B B A+ = + Commutative property of addition 
b. An m n× A matrix added to any m n× zero matrix is equal to itself, i.e., 
A Am n m n m n× × ×+ =0 
c. An m n× A matrix added to the negative of an m n× A matrix is equal to the m n× zero 
matrix, i.e., 
( )A Am n m n m n× × ×+ − = 0 and − + =× × ×A Am n m n m n0 
Example 3.2-2: Given A = −





1 3
2 5
 find. 
a. ( )A A+ − = b. A At+ = c. A I+ = d. A + 0 = 
Solutions: 
a. ( )A A+ − = 1 3
2 5
1 3
2 5
−




 −
−




 = 
1 3
2 5
1 3
2 5
−




 +
−
− −





 = 
1 1 3 3
2 2 5 5
− − +
− −





 = 
0 0
0 0





 = 02 2× 
 
b. A At+ = 
1 3
2 5
1 2
3 5
−




 + −





 = 
1 1 3 2
2 3 5 5
+ − +
− +





 = 
2 1
1 10
−
−





 
 
c. A I+ = 
1 3
2 5
1 0
0 1
−




 +





 = 
1 1 3 0
2 0 5 1
+ − +
+ +





 = 
2 3
2 6
−




 
 
d. A + 0 = 
1 3
2 5
0 0
0 0
−




 +





 = 
1 0 3 0
2 0 5 0
+ − +
+ +





 = 
1 3
2 5
−




 
Example 3.2-3: Given A = 





1 2
3 5
, B =
−






2 4
3 1
, and C = 





1 3
0 5
 find. 
a. A B+ = b. B A+ = c. ( )A B C+ + = d. ( )A B C+ + = 
e. ( )A B Ct t+ + = f. A B Ct t+ + = g. ( )A A C+ − + = 
Solutions: 
a. A B+ = 
1 2
3 5
2 4
3 1





 + −





 = 
1 2 2 4
3 3 5 1
+ +
− +





 = 
3 6
0 6





 
Mastering Algebra - Advanced Level 3.2 Matrix Operations 
Hamilton Education Guides 175 
 
b. B A+ = 
2 4
3 1
1 2
3 5−





 +





 = 
2 1 4 2
3 3 1 5
+ +
− + +





 = 
3 6
0 6





 
Note that since matrix addition is commutative, the results in parts a and b are the same. 
c. ( )A B C+ + = 1 2
3 5
2 4
3 1
1 3
0 5





 + −





 +











 = 
1 2
3 5
2 1 4 3
3 0 1 5





 +
+ +
− + +





 = 
1 2
3 5
3 7
3 6





 + −





 
 
 = 
1 3 2 7
3 3 5 6
+ +
− +





 = 
4 9
0 11





 
 
d. ( )A B C+ + = 1 2
3 5
2 4
3 1
1 3
0 5





 + −











 +





 = 
3 6
0 6
1 3
0 5





 +





 = 
3 1 6 3
0 0 6 5
+ +
+ +





 = 
4 9
0 11





 
Note that since matrix addition is associative, the results in parts c and d are the same. 
e. ( )A B Ct t+ + = 1 23 5
2 3
4 1
1 0
3 5





 +
−




 +











 = 
1 2
3 5
2 1 3 0
4 3 1 5





 +
+ − +
+ +





 = 
1 2
3 5
3 3
7 6





 +
−




 
 
 = 
1 3 2 3
3 7 5 6
+ −
+ +





 = 
4 1
10 11
−




 
 
f. A B Ct t+ + = 
1 3
2 5
2 4
3 1
1 0
3 5





 + −





 +





 = 
1 2 3 4
2 3 5 1
1 0
3 5
+ +
− +





 +





 = 
3 7
1 6
1 0
3 5−





 +





 = 
4 7
2 11





 
 
g. ( )A A C+ − + = 1 2
3 5
1 2
3 5
1 3
0 5





 −





 +





 = 
1 1 2 2
3 3 5 5
1 3
0 5
− −
− −





 +





 = 
0 0
0 0
1 3
0 5





 +





 = 
1 3
0 5





 
Example 3.2-4: Given A =
−










1 2 3
4 3 1
1 1 0
 and B =
− −
−










1 3 1
4 5 6
1 0 3
 find. 
a. A B+ = b. B A+ = c. ( )A B I+ + = d. ( )A B Bt+ + = 
Solutions: 
a. A B+ = 
1 2 3
4 3 1
1 1 0
1 3 1
4 5 6
1 0 3−










+
− −
−










 = 
1 1 2 3 3 1
4 4 3 5 1 6
1 1 1 0 0 3
− − +
+ + +
+ − + −










 = 
0 1 4
8 8 7
2 1 3
−
− −










 
 
b. B A+ = 
− −
−










+
−










1 3 1
4 5 6
1 0 3
1 2 3
4 3 1
1 1 0
 = 
− + − + +
+ + +
+ − − +










1 1 3 2 1 3
4 4 5 3 6 1
1 1 0 1 3 0
 = 
0 1 4
8 8 7
2 1 3
−
− −










 
Mastering Algebra - Advanced Level 3.2 Matrix Operations 
Hamilton Education Guides 176 
c. ( )A B I+ + = 
1 2 3
4 3 1
1 1 0
1 3 1
4 5 6
1 0 3
1 0 0
0 1 0
0 0 1−










+
− −
−




















+










 = 
0 1 4
8 8 7
2 1 3
1 0 0
0 1 0
0 0 1
−
− −










+










 = 
0 1 1 0 4 0
8 0 8 1 7 0
2 0 1 0 3 1
+ − + +
+ + +
+ − + − +










 
 
 = 
1 1 4
8 9 7
2 1 2
−
− −










 
 
d. ( )A B Bt+ + = 
1 2 3
4 3 1
1 1 0
1 3 1
4 5 6
1 0 3
1 3 1
4 5 6
1 0 3−










+
− −
−




















+
− −
−










t
 = 
0 1 4
8 8 7
2 1 3
1 3 1
4 5 6
1 0 3
−
− −










+
− −
−










t
 
 
 = 
0 8 2
1 8 1
4 7 3
1 3 1
4 5 6
1 0 3
− −
−










+
− −
−










 = 
0 1 8 3 2 1
1 4 8 5 1 6
4 1 7 0 3 3
− − +
− + + − +
+ + − −










 = 
−









1 5 3
3 13 5
5 7 0
 
Example 3.2-5: Solve the following matrix operations. 
a. 2 3
1 0
2 3
1 0
1 3
2 8





 +











 −





 = b. 
2 3
5 1
0 2
8 1
3 5
1 2
1 1
2 3
3 5










−
− −










−
−
−




















 = 
Solutions: 
a. 2 3
1 0
2 3
1 0
1 3
2 8





 +











 −





 = 
2 2 3 3
1 1 0 0
1 3
2 8
+ +
+ +





 −





 = 
4 6
2 0
1 3
2 8





 −





 = 
4 1 6 3
2 2 0 8
− −
− −





 = 
3 3
0 8−





 
 
b. 
2 3
5 1
0 2
8 1
3 5
1 2
1 1
2 3
3 5










−
− −










−
−
−




















 = 
2 3
5 1
0 2
8 1 1 1
3 2 5 3
1 3 2 5










−
− +
− −
− + − −










 = 
2 3
5 1
0 2
7 2
1 2
2 7










−
−










 = 
2 7 3 2
5 1 1 2
0 2 2 7
− −
− −
− +










 
 
 = 
−
−
−










5 1
4 1
2 9
 
Section 3.2 Case I Practice Problems - Matrix Addition and Subtraction 
1. Add or subtract the following matrices. 
a. 1 3 5
3 1 2
1 5 2
0 6 3−





 +
− −
−





 = b. 
1 8
3 5
1 3
2 4
−
−





 −
−
−





 = c. 
10 3 1
1 5 6
2 3 8
1 7 9
2 3 0
1 5 3
−








+ −
−










 = 
d. 
2 3
1 5
3 1
1 8
3 0
1 2−










−
−










 = e. [ ] [ ]1 3 6 0 5 3− − = f. 1 53 8
0 0
0 0
−
−





 +





 = 
Mastering Algebra - Advanced Level 3.2 Matrix Operations 
Hamilton Education Guides 177 
 
g. 
6 3 1
1 5 0
0 0 1
1 0 0
0 1 0
0 0 1
−









+










 = h. 
1 5
2 3
1 0
1 5
2 3
1 0
−
−










+
− −
−










 = 
2. Given A = −
−










1 3 5
2 1 0
1 2 4
 and B =
−
−










1 0 1
2 3 0
1 5 3
 find. 
a. A Bt+ = b. A Bt t+ = c. ( )A B Bt+ − = d. ( )A B Bt− + = 
e. ( ) ( )A B A Bt t+ − + = f. ( )A B At− − = g. 2 3A B− = 
3. Given [ ]A = −1 0 3 , [ ]B = 3 1 5 , C = −





1 5 1
2 0 3
, and D =
−






10 8 7
3 5 2
 perform the following 
operations, if possible. 
a. A C+ = b. A B− = c. ( ) tDDC −+ = 
4. Add or subtract the following matrices. 
a. 1 2
3 0
1 3
5 2
2 3
1 6





 − −











 +
−





 = b. 
1 3
5 0
1 1
2 3
1 0
1 3
1 5
1 2
4 3−










− −










− −
−




















 = 
c. 
1 3 5
2 1 3
1 4 0
1 5 0
0 1 1
1 2 3
1 0 0
0 1 0
0 0 1
−
−










+ −










+




















 = d. 2 3
1 0
2 3
1 0
3 5
1 2−





 +
− −










 −
−





 = 
5. Given A =
−
−










2 3 1
5 0 2
8 3 1
 show that [ ]A At t = . 
6. Given A = 





1 2
0 3
 and B = −
−






3 5
3 2
 show that [ ]A B A Bt t t+ = + . 
7. Given the equal matrices solve for the unknowns. 
a. 
u v
z
v
w x
y
−











=
+
+
−










1 3
8 6 16
0 6
3
5 2 3 9
1 6 2
0 2 3 3
 b. 3 5
3 8
3 6
5 4
6 11
2 4
x
y −





 + −





 = − −





 
c. 3 5
3 8
7 1
2 2
4 4
5 10−





 − −





 =
−
−





x
 d. 
x y
y
z
z
z
0
0
3
0
2
3
6
3
9










+










+ −










=










 
Mastering Algebra - Advanced Level 3.2 Matrix Operations 
Hamilton Education Guides 178 
Case II Matrix Multiplication 
Two matrices A and B can be multiplied by one another only if the number of columns in the A 
matrix equals the number of rows in the B matrix, i.e., 
A B Cno of rows in A no of cols in A no of rows in B no of cols in B no of rows in A no of cols in B. . . . . . . . .× × ×× = 
 
 no of columns in A must match the no of rows in B. . 
For example, we can only multiply a 23× ( )23×A , a 34× ( )34×A or a 53× ( )53×A matrix by a 22× 
( )22×B , 53× ( )53×B or a 75× ( )75×B matrix, respectively. The order of the product matrices would 
then be equal to: 
232223 ××× =× CBA , 545334 ××× =× CBA , and 737553 ××× =× CBA 
Multiplication of matrices requires calculating sums of products. In general, a 32× ( )32×A matrix 
is multiplied by a 33× ( )33×B matrix in the following way: 






232221
131211
aaa
aaa
×










333231
232221
131211
bbb
bbb
bbb
 = 





232221
131211
ccc
ccc
 
where the entries for the C matrix are calculated in the following way: 
c a b a b a b11 11 11 12 21 13 31= ⋅ + ⋅ + ⋅ c a b a b a b12 11 12 12 22 13 32= ⋅ + ⋅ + ⋅ c a b a b a b13 11 13 12 23 13 33= ⋅ + ⋅ + ⋅ 
c a b a b a b21 21 11 22 21 23 31= ⋅ + ⋅ + ⋅ c a b a b a b22 21 12 22 22 23 32= ⋅ + ⋅ + ⋅ c a b a b a b23 21 13 22 23 23 33= ⋅ + ⋅ + ⋅ 
The following examples show how two matrices are multiplied by one another: 
Example 3.2-6: Multiply the following matrices. 
a. [ ]










1
0
2
321 = b. 2 4
3 1
1 2
0 1−





 −





 = c. 
1 0 2
3 1 4
1 2
1 3
2 5
−





 −










 = 
Solutions: 
a. [ ]










1
0
2
321 = ( ) ( ) ( )130221 ×+×+× = 302 ++ = 5 
From the above example it should be clear that in general the product of a matrix with one row 
and n columns ( )nA ×1 multiplied by another matrix with n rows and one column ( )1×nB is always 
a real number - not a matrix, i.e., kBA nn =⋅ ×× 11 . 
b. 
2 4
3 1
1 2
0 1−





 −





 = 
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
2 1 4 0 2 2 4 1
3 1 1 0 3 2 1 1
× + × × + × −
− × + × − × + × −





 = 
2 0 4 4
3 0 6 1
+ −
− + − −





 = 
2 0
3 7− −





 
 
c. 
1 0 2
3 1 4
1 2
1 3
2 5
−





 −










 = 
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
1 1 0 1 2 2 1 2 0 3 2 5
3 1 1 1 4 2 3 2 1 3 4 5
× + × − + × × + × + ×
× + − × − + × × + − × + ×





 = 
1 0 4 2 0 10
3 1 8 6 3 20
+ + + +
+ + − +





 = 
5 12
12 23





 
Mastering Algebra - Advanced Level 3.2 Matrix Operations 
Hamilton Education Guides 179 
Properties of Matrix Multiplication 
a. Matrix multiplication is not commutative. This implies that, for most matrices, AB BA≠ . 
Example 3.2-7: Given A =
−






2 3
1 5
 and B = 





1 2
0 1
, show that AB BA≠ . 
Solution: 
AB = 





⋅





− 10
21
51
32 = ( ) ( ) ( ) ( )( ) ( ) ( ) ( )
2 1 3 0 2 2 3 1
1 1 5 0 1 2 5 1
× + × × + ×
− × + × − × + ×





 = 
2 0 4 3
1 0 2 5
+ +
− + − +





 = 
2 7
1 3−





 
 
BA = 





−
⋅





51
32
10
21 = ( ) ( ) ( ) ( )( ) ( ) ( ) ( )
1 2 2 1 1 3 2 5
0 2 1 1 0 3 1 5
× + × − × + ×
× + × − × + ×





 = 
2 2 3 10
0 1 0 5
− +
− +





 = 
0 13
1 5−





 
b. If A , B , and C are square matrices and a and b are real number, then 
1. ( ) ( )AB C A BC= 
Example 3.2-8: Given A =
−






0 1
1 3
 , B = −





1 2
5 1
, and C = −
−






3 4
1 1
 show that ( ) ( )AB C A BC= . 
Solution: 
( )AB C = 












−
−
⋅




−
⋅





− 11
43
15
21
31
10 = ( ) ( ) ( ) ( )( ) ( ) ( ) ( )
0 1 1 5 0 2 1 1
1 1 3 5 1 2 3 1
3 4
1 1
× − + × × + ×
− × − + × − × + ×











 ⋅
−
−





 
 
 = 
0 5 0 1
1 15 2 3
3 4
1 1
+ +
+ − +





 ⋅
−
−





 = 
5 1
16 1
3 4
1 1





 ⋅
−
−





 = 
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
5 3 1 1 5 4 1 1
16 3 1 1 16 4 1 1
× − + × − × + ×
× − + × − × + ×





 
 
 = 
− − +
− − +






15 1 20 1
48 1 64 1
 = 
−
−






16 21
49 65 
 
( )BCA = 












−
−
⋅




−
⋅





− 11
43
15
21
31
10 = ( ) ( ) ( ) ( )( ) ( ) ( ) ( )
0 1
1 3
1 3 2 1 1 4 2 1
5 3 1 1 5 4 1 1−





 ⋅
− × − + × − − × + ×
× − + × − × + ×











 
 
 = 
0 1
1 3
3 2 4 2
15 1 20 1−





 ⋅
− − +
− − +





 = 
0 1
1 3
1 2
16 21−





 ⋅
−
−





 = 
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
0 1 1 16 0 2 1 21
1 1 3 16 1 2 3 21
× + × − × − + ×
− × + × − − × − + ×





 
 
 = 
0 16 0 21
1 48 2 63
− +
− − +





 = 
−
−






16 21
49 65
 
2. ( )A B C AB AC+ = + 
Example 3.2-9: Given A = −
−






4 3
1 3
, B = −





1 2
4 1
, and C = 





0 5
3 0
 show that ( )A B C AB AC+ = + . 
Solution: 
( )CBA + = 












+




 −






−
−
03
50
14
21
31
34 = 
−
−





 ⋅
+ − +
+ +






4 3
1 3
1 0 2 5
4 3 1 0
 = 
−
−





 ⋅






4 3
1 3
1 3
7 1
 
 = ( ) ( )( ) ( )( ) ( ) ( ) ( )
− × + × − × + ×
× + − × × + − ×






4 1 3 7 4 3 3 1
1 1 3 7 1 3 3 1
 = 
− + − +
− −






4 21 12 3
1 21 3 3
 = 
17 9
20 0
−
−






Mastering Algebra - Advanced Level 3.2 Matrix Operations 
Hamilton Education Guides 180 
AB = 




 −
⋅





−
−
14
21
31
34 = ( ) ( ) ( ) ( )( ) ( ) ( ) ( )
− × + × − × − + ×
× + − × × − + − ×






4 1 3 4 4 2 3 1
1 1 3 4 1 2 3 1
 = 
− + +
− − −






4 12 8 3
1 12 2 3
 = 
8 11
11 5− −





 
 
AC = 





⋅





−
−
03
50
31
34 = ( ) ( ) ( ) ( )( ) ( ) ( ) ( )
− × + × − × + ×
× + − × × + − ×






4 0 3 3 4 5 3 0
1 0 3 3 1 5 3 0
 = 
0 9 20 0
0 9 5 0
+ − +
− +





 = 
9 20
9 5
−
−





 
 
ACAB + = 





−
−
+





−− 59
209
511
118 = 
8 9 11 20
11 9 5 5
+ −
− − − +





 = 
17 9
20 0
−
−





 
3. ( )B C A BA CA+ = + 
Example 3.2-10: Given A = − −





5 2
0 2
, B =
−






1 0
0 1
, and C =
−






2 0
3 4
 show that ( )B C A BA CA+ = + . 
Solution: 
( )ACB + = 




 −−














−
+





− 20
25
43
02
10
01 = 
1 2 0 0
0 3 1 4
5 2
0 2
+ +
+ − −






− −




 = 
3 0
3 5
5 2
0 2−






− −




 
 
 = ( ) ( ) ( ) ( )( ) ( ) ( ) ( )
3 5 0 0 3 2 0 2
3 5 5 0 3 2 5 2
× − + × × − + ×
× − + − × × − + − ×





 = 
− + − +
− + − −






15 0 6 0
15 0 6 10
 = 
− −
− −






15 6
15 16
 
 
BA = 




 −−
⋅





− 20
25
10
01 = ( ) ( ) ( ) ( )( ) ( ) ( ) ( )
1 5 0 0 1 2 0 2
0 5 1 0 0 2 1 2
× − + × × − + ×
× − + − × × − + − ×





 = 
− + − +
+ −






5 0 2 0
0 0 0 2
 = 
− −
−






5 2
0 2 
 
CA = 




 −−
⋅





− 20
25
43
02 = ( ) ( ) ( ) ( )( ) ( ) ( ) ( )
2 5 0 0 2 2 0 2
3 5 4 0 3 2 4 2
× − + × × − + ×
× − + − × × − + − ×





 = 
− + − +
− + − −






10 0 4 0
15 0 6 8
 = 
− −
− −






10 4
15 14 
 
CABA+ = 





−−
−−
+





−
−−
1415
410
20
25 = 
− − − −
− − −






5 10 2 4
0 15 2 14
 = 
− −
− −






15 6
15 16 
4. ( ) ( ) ( )a AB aA B A aB= = 
Example 3.2-11: Given A =
−
−










1 0 1
2 3 1
0 5 2
 , B =
− −
−
−










3 2 1
1 0 1
0 3 6
, and a = −5 show that the equalities 
( ) ( ) ( )a AB aA B A aB= = are true. 
Solution: 
AB = 
1 0 1 3 2 1
2 3 1 1 0 1
0 5 2 0 3 6
− − −
− ⋅ −
−
   
   
   
      
 = 
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
1 3 0 1 1 0 1 2 0 0 1 3 1 1 0 1 1 6
2 3 3 1 1 0 2 2 3 0 1 3 2 1 3 1 1 6
0 3 5 1 2 0 0 2 5 0 2 3 0 1 5 1 2 6
× − + × − + − × × + × + − × × − + × + − × −
× − + × − + − × × + × + − × × − + × + − × −
× − + × − + × × + × + × × − + × + × −










 
 
= 
− + + + − − + +
− − + + − − + +
− + + + + −










3 0 0 2 0 3 1 0 6
6 3 0 4 0 3 2 3 6
0 5 0 0 0 6 0 5 12
 = 
− −
−
− −










3 1 5
9 1 7
5 6 7
. thus, ( )5 AB− = 










−−
−
−−
−
765
719
513
.5 = 
15 5 25
45 5 35
25 30 35
−
− −
−










 
Mastering Algebra - Advanced Level 3.2 Matrix Operations 
Hamilton Education Guides 181 
( )− = − ⋅
−
−




















⋅
− −
−
−










5 5
1 0 1
2 3 1
0 5 2
3 2 1
1 0 1
0 3 6
A B = 
− × × − − × −
− × − × − × −
− − × − − ×










⋅
− −
−
−










5 1 0 5 1 5
5 2 5 3 5 1
0 5 5 5 5 2
3 2 1
1 0 1
0 3 6
 = 
−
− −
− −










⋅
− −
−
−










5 0 5
10 15 5
0 25 10
3 2 1
1 0 1
0 3 6
 
 
= 
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
− × − + × − + × − × + × + × − × − + × + × −
− × − + − × − + × − × + − × + × − × − + − × + × −
× − + − × − + − × × + − × + − × × − + − × + − × −










5 3 0 1 5 0 5 2 0 0 5 3 5 1 0 1 5 6
10 3 15 1 5 0 10 2 15 0 5 3 10 1 15 1 5 6
0 3 25 1 10 0 0 2 25 0 10 3 0 1 25 1 10 6
 
 
= 
15 0 0 10 0 15 5 0 30
30 15 0 20 0 15 10 15 30
0 25 0 0 0 30 0 25 60
+ + − + + + −
+ + − + + − −
+ + + − − +










 = 
15 5 25
45 5 35
25 30 35
−
− −
−










 therefore ( ) ( )− = −5 5AB A B 
 
( )




















−
−
−−
⋅−










−
−
=−
630
101
123
5
250
132
101
5BA = 
1 0 1
2 3 1
0 5 2
5 3 5 2 5 1
5 1 5 0 5 1
5 0 5 3 5 6
−
−










− × − − × − × −
− × − − × − ×
− × − × − × −










 = 
1 0 1
2 3 1
0 5 2
15 10 5
5 0 5
0 15 30
−
−










−
−
−










 
 
= 
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
1 15 0 5 1 0 1 10 0 0 1 15 1 5 0 5 1 30
2 15 3 5 1 0 2 10 3 0 1 15 2 5 3 5 1 30
0 15 5 5 2 0 0 10 5 0 2 15 0 5 5 5 2 30
× + × + − × × − + × + − × − × + × − + − ×
× + × + − × × − + × + − × − × + × − + − ×
× + × + × × − + × + × − × + × − + ×










 
 
= 
15 0 0 10 0 15 5 0 30
30 15 0 20 0 15 10 15 30
0 25 0 0 0 30 0 25 60
+ + − + + + −
+ + − + + − −
+ + + − − +










 = 
15 5 25
45 5 35
25 30 35
−
− −
−










 therefore ( ) ( ) ( )− = − = −5 5 5AB A B A B 
5. ( ) ( )a bA ab A= 
Example 3.2-12: Given A = −
−






5 4
1 5
, a = 3 , and b = −1 show that ( ) ( )a bA ab A= . 
Solution: 
( )bAa = 












−
−
⋅−⋅
51
45
13 = 3
5 1 4 1
1 1 5 1
⋅
− × − × −
− × − × −





 = 3
5 4
1 5
⋅
−
−





 = 
5 3 4 3
1 3 5 3
× − ×
× − ×





 = 
15 12
3 15
−
−





 
 
( )Aab = ( ) 





−
−
−×
51
45
13 = − ⋅
−
−





3
5 4
1 5
 = 
− × − × −
− × − × −






5 3 4 3
1 3 5 3
 = 
15 12
3 15
−
−





 
6. ( )a A B aA aB+ = + 
Example 3.2-13: Given A = −
−






6 10
2 3
 , B =
−






3 2
1 0
, and a = 4 . Show that ( )a A B aA aB+ = + . 
Solution: 
( )BAa + = 












−
+





−
−
⋅
01
23
32
106
4 = 4
6 3 10 2
2 1 3 0
⋅
− + +
− − +





 = 4
3 12
3 3
⋅
−
−





 = 
− × ×
− × ×






3 4 12 4
3 4 3 4
 = 
−
−






12 48
12 12 
 
 
Mastering Algebra - Advanced Level 3.2 Matrix Operations 
Hamilton Education Guides 182 
aBaA+ = 





−
⋅+





−
−
⋅
01
23
4
32
106
4 = 
− × ×
− × ×





 +
× ×
− × ×






6 4 10 4
2 4 3 4
3 4 2 4
1 4 0 4
 = 
−
−





 + −






24 40
8 12
12 8
4 0
 
= 
− + +
− − +






24 12 40 8
8 4 12 0
 = 
−
−






12 48
12 12
 
7. The following are additional properties of matrix multiplication. 
 a. One multiplied by a matrix A is always equal to A , i.e., 1⋅ =A A . 
 b. Minus one multiplied by a matrix A is equal to −A , i.e., − ⋅ = −1 A A . 
 c. Zero multiplied by a matrix A is equal to the zero matrix, i.e., 0 0⋅ =A . 
 d. A constant a multiplied by the zero matrix is equal to the zero matrix, i.e., a ⋅ =0 0 . 
 e. Matrix A multiplied by the identity matrix is equal to the A matrix, i.e., I A A I A⋅ = ⋅ =
. 
 f. A constant a multiplied by a matrix A is equal to the B matrix, i.e., a A B⋅ = . 
 Note that the entries of the B matrix are obtained by multiplying the constant a with 
 each entries of the matrix A . 
Example 3.2-14: Given A = 





7 3
4 0
, I = 





1 0
0 1
, and a = −6 show that: a. 1⋅ =A A , b. − ⋅ = −1 A A