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CEFET – MG CAMPUS III – LEOPOLDINA
Professor José Eduardo Salgueiro
3ª SÉRIES INTEGRADO - DIURNO MATEMÁTICA
3° BIMESTRE -2013
VALOR: 2 PONTOS
1) Os valores reais a e b, tais que os polinômios:
��– 2��� + (3� + �)�– 3� e � �– (� + 2�)� + 2� sejam divisíveis por (� + 1), são:
��– 2��� + (3� + �)�– 3� � + 1
−�� − ��
−(2� + 1)�� + (3� + �)�– 3�
(2� + 1)�� + (2� + 1)�
(5� + � + 1)� − 3�
−(5� + � + 1)� − (5� + � + 1)
−5� − 4� − 1���������
���� � (�)
�� − (2� + 1)� + (5� + � + 1)
���������������������
����� ��� � (�)
��– (� + 2�)� + 2� � + 1
−�� − ��
−�� − (� + 2�)� + 2�
�� + �
−(� + 2� − 1)� + 2�
−(� + 2� − 1)� + (� + 2� − 1)
3� + 2� − 1���������
����� (�)
�� − � − (� + 2� − 1)
���������������
����� ��� �(�)
−1 1 −2� 3� + � −3�
1 −1 ∙ 1 − 2� −1 ∙ [−(1 + 2�)] + 3� + � −1 ∙ (5� + � + 1) − 3�
−(1 + 2�) 5� + � + 1 −5� − 4� − 1
��(�) = 0
−1 1 0 – (� + 2�) 2�
1 −1 ∙ 1 + 0 −1 ∙ (−1) − (� + 2�) −1 ∙ (1 − � − 2�) + 2�
−1 1 − � − 2� 3� + 2� − 1
��(�) = 0
�
−5� − 4� − 1
3� + 2� − 1(2)
�
−5� − 4� − 1 = 0
6� + 4� − 2 = 0
� = 3 � = −4
Conferindo:
�
�
� �– 2��� + (3� + �)�– 3�
��– 2(3)�� + [3(3) + (−4)]�– 3(−4)
��– 6�� + 5� + 12
�(−1) = (−1)�– 6(−1)� + 5(−1) + 12
�(−1) = −1 − 6 − 5 + 12
�(−1) = 0 ��
�
�
��– (� + 2�)� + 2�
��– [(3) + 2(−4)]� + 2(3)
�� + 5� + 6
�(−1) = (−1)� + 5(−1) + 6
�(−1) = −1 − 5 + 6
�(−1) = 0 ��
2) A divisão de um polinômio P(x) por ��– � resulta no quociente 6�� + 5� + 3 e
resto – 7�. O resto da divisão de �(�) por 2� + 1 é igual a:
�(�) = (��– � ) ∙ (6�� + 5� + 3) + (−7�)
�(�) = 6�� + 5�� + 3�� − 6�� − 5�� − 3� − 7�
�(�) = 6�� − �� − 2� � − 10�
6�� − �� − 2� � − 10� 2� + 1
−6� � − 3��
−4�� − 2� � − 10�
+4�� + 2��
−10�
10� + 5
�������� ∶ 5�����������
����� (�)
3�� − 2�� − 5
���������
����� ��� �(�)
3) Transforme num polinômio reduzido e ordenado segundo as potencias
decrescentes de x o polinômio
�(�) = 4 �� −
1
2
� �
1
4
− �� − 2(2 − �)�
�(�) = 4 �
�
4
− �²−
1
8
+
�
2
� − 2(4 − 4� + �²) = � − 4�²−
1
2
+ 2� − 8 + 8� − 2�²
��� ��� ��: �(�) = −6�²+ 11� −
17
2
4) Calcular A, B � � , de modo que:
�
�
+
�
� − 1
+
�
� − 2
=
18�²− 13� + 2
�³− 3�²+ 2x
SOLUÇÃO:
�(� − 1)(� − 2) + �(�)(� − 2) + �(�)(� − 2)
�(� − 1)(� − 2)
=
18�² − 13� + 2
�³ − 3�²+ 2x
�(�²− 3� + 2) + ��²− 2�� + ��²− ��
�³− 3�²+ 2x
=
18�²− 13� + 2
�³− 3�²+ 2x
��²− 3�� + 2� + ��² − 2�� + ��²− �� = 18�² − 13� + 2
(� + � + �)² + (−3� − 2� − �)� + 2� = 18�²− 13� + 2
�
� + � + � = 18 (1)
−3� − 2� − � = −13 (2)
2� = 2 (3)
�
� + � + � = 18 (1)
−3� − 2� − � = −13 (2)
� = 1 ����� .(3), � � (1) � (2)
�
1 + � + � = 18 (1)
−3(1) − 2� − � = −13 (2)
�
� + � = 17 (1)
−2� − � = −10 (2)
, � ������ (1)� � � (2), � ���� :
−� = 7 ∴ � = −7 ∴ ����� .� � (1), � ���� : −7 + � = 17 � = 24
��� ��� ��: � = 1, � = −7 � � = 24
5) Num polinômio �(�) do 3° Grau, o coeficiente de �³ é 1. Se �(1) = �(2) = 0
� �(3) = 30, ��� ��� � � ��� �� �� �(−1).
SOLUÇÃO:
Se �(�) é do 3° Grau, temos: �(�) = ��³ + ��² + �� + �, � � � � = 1.
�(1) = 1(�)� + �(1)� + �(1) + � ∴ 1 + � + � + � = 0 ∴ � + � + � = −1 (1)
�(2) = 1(2)� + �(2)� + �(2) + � ∴ 8 + 4� + 2� + � = 0 ∴ 4� + 2� + � = −8 (2)
�(3) = 1(3)� + �(3)� + �(3) + � ∴ 27 + 9� + 3� + � = 30 ∴ 9� + 3� + � = 3(3)
�
� = −1 − � − � (1)
4� + 2� + � = −8 (2)
9� + 3� + � = 3 (3)
, ����� .(1)� � (2)�(3), � ���� :
�
4� + 2� − 1 − � − � = −8 (2)
9� + 3� − 1 − � − � = 3 (3)
�
3� + � = −7 (2), ��� �� � � � ����� ��� − 2
8� + 2� = 4 (3)
,
temos:
�
−6� − 2� = 14 (2)
8� + 2� = 4 (3)
, � ������ (2)� � � (3), � ���� : 2� = 19 ∴ � = 9(4);
����� .(4)� � (2), � ���� : −6(9) − 2� = 14 ∴ −54 − 2� = 14 ∴ � = −34(5);
����� .(4), (5)� � � = −1 − � − � (1), � ���� : � = −1 − 9 − (−34) = � = 24(6);
����� . Os valores de �, �, � � �, no polinômio �(�) = ��³+ ��² + �� + �, � ���� :
�(�) = ��³+ ��² + �� + � ∴ �(�) = �³ + 9�²− 34� + 24
�(−1) = (−1)� + 9(−1)� − 34(−1) + 24 ∴ �(−1) = −1 + 9 + 34 + 24
Resposta: �(−1) = 66
6) Determinar o quociente e o resto da divisão de:
�(�) = 2�� − 13�³ + 13�² + 6� − 9 ��� �(�) = �³− 5�²− � + 1
a) Usando o Método da Chave;
b) Usando o Método dos coeficientes a determinar;
SOLUÇÃO:
Usando o método da chave:
2�� − 13�³ + 13�²+ 6� − 9 �³ − 5�²− � + 1
−2�� + 10�³ + 2�²− 2� 2� − 3 �������
����� ��� �
−3�³+ 15�²+ 4� − 9
3�³− 15�²− 3� + 3
��� �.: ���� �(�) = � − 6
Usando o Método dos coeficientes a determinar (Método de Descartes)
�� (�) = �� (�) − �� (�) = 4 − 3 = 1, ��� ã�: �(�) = �� − �
�� (�) < �� (�), ��� ã��� (�) < 3 , � ���� �(�) = ��² + �� + �
Se �(�) ≡ �(�) × �(�) + �(�), � ���� :
2�� − 13�³+ 13�² + 6� − 9 ≡ (�³− 5�²− � + 1)(�� + �) + ��²+ �� + �
2�� − 13�³+ 13�² + 6� − 9
≡ ��� + ��³− 5��³− 5��²− ��² − �� + �� + � + ��²+ �� + �
2�� − 13�³+ 13�² + 6� − 9 ≡ ��� + (� − 5�)�� + (−5� − � + �)�� + (−� + � +
�)� + (� + �)
�
� = 2
� = 2
�
� − 5� = −13
� − 5(2) = −13
� = −3
�
−5� − � + � = 13
−5(−3) − 2 + � = 13
� = 0
�
−� + � + � = 6
−(−3) + 2 + � = 6
� = 1
�
� + � = −9
−3 + � = −9
� = −6
�(�) = �� + � ∴ �(�) = 2� + (−3)
�(�) = ��²+ �� + � ∴ �(�) = 0�²+ � − 6
RESPOSTA:
�(�) = 2� − 3
�(�) = � − 6
7) Um Polinômio�(�) dividido por (� − 1) �á ����� 2, ��� (� − 2) �á ����� 1
� ��� (� − 3) �á ����� − 4. Calcular o resto da divisão de
�(�) ��� (� − 1)(� − 2)(� − 3)
�(1) = 2 ∴ �(2) = 1 ∴ �(3) = −4
�� → �� (�) < �� (�), ��� ã��� (�) < 3 , � ���� �(�) = ��²+ �� + �
Se �(�) ≡ �(�) × �(�) + �(�), � ���� :
�(�) ≡ (� − 1)(� − 2)(� − 3) × �(�) + ��² + �� + �
���� �(1) = 2, � ���� �(1) ≡ (1 − 1)(1 − 2)(1 − 3) × �(�) + �(1)� + �(1) + � = 2
�(1) ≡ 0 × �(�) + �(1)� + �(1) + � = 2 ∴ � + � + � = 2 (1)
���� �(2) = 1, � ���� �(2) ≡ (2 − 1)(2 − 2)(2 − 3) × �(�) + �(2)� + �(2) + � = 1
�(2) ≡ 0 × �(�) + �(2)� + �(2) + � = 1 ∴ 4� + 2� + � = 1 (2)
���� �(3) = −4, � ���� �(3) ≡ (3 − 1)(3 − 2)(3 − 3) × �(�) + �(3)� + �(3) + � = −4
�(3) ≡ 0 × �(�) + �(3)� + �(3) + � = −4 ∴ 9� + 3� + � = −4 (3)
Resolvendo o sistema com (1),(2)e(3):
�
� + � + � = 2 (1)
4� + 2� + � = 1 (2)
9� + 3� + � = −4(3)
�
� = 2 − � − � ����� .(1)� � (2)�(3)
4� + 2� + 2 − � − � = 1 (2)
9� + 3� + 2 − � − � = −4 (3)
�
� = 2 − � − � (1)
3� + � = −1 × (−2)(2)
8� + 2� = −6 (3)
�
−6� − 2� = 2 (2)
8� + 2� = −6 (3)
, � ������ (2)� � � (3), � ���� : 2� = −4 ∴ � = −2 (4)
����� .(4)� � (2), � ���� − 6(−2) − 2� = 2 ∴� = 5 (5)
����� .(4), (5)� � (1), � ���� � = 2 − (−2) − 5 ∴ � = −1 (6)
����� .�� ��� ���� �� �, � � � � � �(�) = ��²+ �� + � ∴ �(�) = −2�²+ 5� − 1
8) Determinar o quociente e o resto da divisão de:
�(�) = �� − 4�³+ 4�²+ 7 ��� �(�) = � − 2
Usando o Método da Chave;
Usando o Método dos coeficientes a determinar;
Usando o Método de Briot-Ruffini
Usando o método da chave:
�� − 4�³+ 4�²+ 7 � − 2
−�� + 2�³ �³− 2�² �������
����� ��� � �(�)
−2�³+ 4�²+ 7
2�³ − 4�²
��� �.: ���� �(�) = 7
Usando o Método dos coeficientes a determinar (Método de Descartes)
�� (�) = �� (�) − �� (�) = 4 − 1 = 3, ��� ã�: �(�) = ��³+ ��²+ �� + �
�� (�) < �� (�), ��� �� (�) < 1 , � ���� : �(�) = �
Se �(�) ≡ �(�) × �(�) + �(�), � ���� :
�� − 4�³+ 4�²+ 7 ≡ (��³+ ��²+ �� + �)(� − 2) + �
�� − 4�³+ 4�²+ 7 ≡ ��� + ��³+ ��²+ �� − 2��³ − 2��² − 2�� − 2� + �
�� − 4�³+ 4�²+ 7 ≡ ��� + (� − 2�)�� + (� − 2�)�� + (� − 2�)� + (−2� + �)
⎩
⎪
⎨
⎪
⎧
� = 1
� − 2� = −4
� − 2� = 4
� − 2� = 0
−2� + � = 7
⎩
⎪
⎨
⎪
⎧
� = 1
� − 2(1) = −4
� − 2� = 4
� − 2� = 0
−2� + � = 7 ⎩
⎪
⎨
⎪
⎧
� = 1
� = −2
� − 2(−2) = 4
� − 2� = 0
−2� + � = 7 ⎩
⎪
⎨
⎪
⎧
� = 1
� = −2
� = 0
� − 2(0) = 0
−2� + � = 7 ⎩
⎪
⎨
⎪
⎧
� = 1
� = −2
� = 0
� = 0
−2(0) + � = 7 ⎩
⎪
⎨
⎪
⎧
� = 1
� = −2
� = 0
� = 0
� = 7
Se �(�) = ��� + ��� + �� + �, � ���� :
�(�) = 1�³− 2�²+ 0� + 0 �(�) = �³− 2�² �� �(�) = �, � ���� : �(�) = 7
C) Usando o Método de Briot-Ruffini
a Coeficientes do dividendo �(�) = �� − 2��
�(�) = 7
2 1 -4 4 7
1 2(1)-4 2(-2)+4 2(0)+7
1 -2 0 7
Coeficientes do quociente Resto
9) Determinar o quociente e o resto da divisão de: P(�) = 4�� + 6�³− 1 por (� + 2)
Usando o Método da Chave;
Usando o Método dos coeficientes a determinar;
Usando o Método de Briot-Ruffini
SOLUÇÃO:
Usando o método da chave:
4�� + 6�³− 1 � + 2
−4�� − 8�³ 4�³− 2�²+ 4� − 8
−2�³ − 1
2�³ + 4�²
4�² − 1 ����� ��� � �(�)
−4�²− 8�
−8� − 1
8� + 16
���� � �(�) 15
Usando o Método dos coeficientes a determinar (Método de Descartes)
�� (�) = �� (�) − �� (�) = 4 − 1 = 3, ��� ã�: �(�) = ��³+ ��²+ �� + �
�� (�) < �� (�), ��� �� (�) < 1 , � ���� : �(�) = �
Se �(�) ≡ �(�) × �(�) + �(�), � ���� :
4�� + 6�� − 1 ≡ (��� + ��� + �� + �)(� + 2) + �
4�� + 6�� − 1 ≡ ��� + ��³+ ��²+ �� + 2��³+ 2��²+ 2�� + 2� + �
4�� + 6�� − 1 ≡ ��� + (� + 2�)�� + (� + 2�)�� + (� + 2�)� + (2� + �)
⎩
⎪
⎨
⎪
⎧
� = 4
� + 2� = 6
� + 2� = 0
� + 2� = 0
2� + � = −1
⎩
⎪
⎨
⎪
⎧
� = 4
� + 2(4) = 6
� + 2� = 0
� + 2� = 0
2� + � = −1 ⎩
⎪
⎨
⎪
⎧
� = 4
� = −2
� + 2(−2) = 0
� + 2� = 0
2� + � = −1 ⎩
⎪
⎨
⎪
⎧
� = 4
� = −2
� = 4
� + 2(4) = 0
−2� + � = −1 ⎩
⎪
⎨
⎪
⎧
� = 4
� = −2
� = 4
� = −8
2(−8) + � = −1 ⎩
⎪
⎨
⎪
⎧
� = 4
� = −2
� = 4
� = −8
� = 15
Se �(�) = ��� + ��� + �� + �, � ���� :
�(�) = 4�³− 2�²+ 4� − 8
�� �(�) = �, � ���� : �(�) = 15
RESPOSTA: �(�) = 4�³− 2�²+ 4� − 8 � �(�) = 15
c) Usando o Método de Briot-Ruffini
a Coeficientes do dividendo
-2 4 6 0 0 -1
4 −2(4) + 6 −2(−2) + 0 −2(4) −2(−8) − 1
4 -2 4 -8 15
Coeficientes do quociente Resto
���� : �(�) = ��� + ��� + �� + �, � ���� : �(�) = 4�³− 2�² + 4� − 8
���� : �(�) = � �(�) = 15
10) Determine o polinômio do 3° grau, que se anula para � = 1 e que, dividido por
� + 1 , � − 2 � � + 2 , apresenta resto igual a 6.
SOLUÇÃO:
Dados: �(1) = 0 ∴ �(−1) = 6 ∴ �(−2) = 6 ∴ �(2) = 6
�� �(�) � � � ���� 3, � ���� �(�) = ��³ + ��² + �� + �
�(1) = 0 ∴ �(1) = �(1)� + �(1)� + �(1) + � ∴ � + � + � + � = 0
�(−1) = 6 ∴ �(−1) = �(−1)� + �(−1)� + �(−1) + � ∴ −� + � − � + � = 6
�(−2) = 6 ∴ �(−2) = �(−2)� + �(−2)� + �(−2) + � ∴ −8� + 4� − 2� + � = 6
�(2) = 6 ∴ �(2) = �(2)� + �(2)� + �(2) + � ∴ 8� + 4� + 2� + � = 6
�
� + � + � + � = 0
−� + � − � + � = 6
+ �
−8� + 4� − 2� + � = 6
8� + 4� + 2� + � = 6
+
2� + 2� = 6 ∴ 8� + 2� = 12(÷ 2) ∴ 4� + � = 6
�
2� + 2� = 6
4� + � = 6
�
−� − � = −3
4� + 2� = 6
∴ 3� = 3 � = 1 ∴ 4(1) + � = 6 ∴ � = 2
�
� + 1 + � + 2 = 0
−8� + 4(1) − 2� + 2 = 6
�
� + � = −3
−8� − 2� = 0
�
8� + 8� = −24
−8� − 2� = 0
�
6� = −24
� = −4
�
� + (−4) = −3
� = 1
�(�) = ��³+ ��² + �� + � ∴ ������� � → �(�) = �³+ �²− 4� + 2
11) Determine m e n de modo que o resto da divisão do polinômio
�(�) = � � − ��� + � , ��� ℎ(�) = �³ + 3�² , � ��ℎ� ����� ���� � � 5.
� ���ÇÃ�:
�� (�) = �� (�) − �� (ℎ) = 5 − 3 = 2, ��� ã�: �(�) = ��²+ �� + � � �(�) = 5
Se �(�) ≡ ℎ(�) × �(�) + �(�), � ���� :
�� − ��� + � = (�³+ 3�² )(��²+ �� + �) + 5
�� − ��� + � = ��� + ��� + ��� + 3��� + 3��� + 3��²+ 5
�� − ��� + � = ��� + (� + 3�)�� + (� + 3�)�� + 3��� + 5
|� = 1 �
� + 3� = 0
� + 3(1) = 0
� = −3
�
� + 3� = −�
� + 3(−3) = −�
0 − 9 = −�
� = 9
�
3� = 0
� = 0
�
� = 5
12) Determine o quociente e o resto da divisão do polinômio
�(�) = ���� + � + 1 ��� �²− 1
���� + � + 1 �� − 1
−(���� − � ��)
��� + ⋯ + � + 1
−(��� − ���)
� �� + ⋯ + � + 1
−(��� − ���) + � + 1
… … … .… … … … … …
�� + ⋯ + � + 1
−(�� − ��)
�� + � + 1
−(�� − 1)
� + 2���
���� �
��� + ��� + ��� + ⋯ + �� + 1���������������������
����� ��� �(�)
13) Decomponha o polinômio �(�) = �� + 5�� + 6� � − 2�� − 7� − 3 = 0, sabendo
que −1 é ��� � ������ da equação.
-1 1 5 6 -2 -7 -3
-1
1
1
-1(1)+5
4
-1(4)+6
2
-1(2)-2
-4
-1(-4)-7
-3
-1(-3)-3
0
1
1
-1(1)+4
3
-1(3)+2
-1
-1(-1)-4
-3
-1(-3)-3
0
�� = 0
-1 1
1
-1(1)+3
2
-1(2)-1
-3
-1(-3)-3
0
�� = 0
�(�) = �� + 2� − 3 = 0 �� = 0
� =
−2 ± �2� − 4(1)(−3)
2(1)
∴ � =
−2 ± 4
2
∴ � � = −3 ∴ ��� = 1
��� ��� �� → �(�) = (� + 1)� × (� + 3) × (� − 1)
14) Determinar o quociente e o resto da divisão de: �(�) = �� − 4�³ + 4�²+ 7
��� �(�) = � − 2.
Usando o Método da Chave;
Usando o Método dos coeficientes a determinar;
Usando o Método de Briot-Ruffini
SOLUÇÃO:
a)Usando o método da chave:
�� − 4�³+ 4�²+ 7 � − 2
−�� + 2�³
−2�³+ 4�²+ 7
2�³− 4�²
7⏟
���� �
�³− 2�²
����� ��� �(�)
b)Usando o Método dos coeficientes a determinar (Método de Descartes)
�� (�) = �� (�) − �� (�) = 4 − 1 = 3, ��� ã�: �(�) = ��³+ ��²+ �� + �
�� (�) < �� (�), ��� �� (�) < 1 , � ���� : �(�) = �
Se �(�) ≡ �(�) × �(�) + �(�), � ���� :
�� − 4�³+ 4�²+ 7 ≡ (��³+ ��²+ �� + �)(� − 2) + �
�� − 4�³+ 4�²+ 7 ≡ ��� + ��³+ ��²+ �� − 2��³ − 2��² −2�� − 2� + �
�� − 4�³+ 4�²+ 7 ≡ ��� + (� − 2�)�� + (� − 2�)�� + (� − 2�)� + (−2� + �)
⎩
⎪
⎨
⎪
⎧
� = 1
� − 2� = −4
� − 2� = 4
� − 2� = 0
−2� + � = 7
⎩
⎪
⎨
⎪
⎧
� = 1
� − 2(1) = −4
� − 2� = 4
� − 2� = 0
−2� + � = 7 ⎩
⎪
⎨
⎪
⎧
� = 1
� = −2
� − 2(−2) = 4
� − 2� = 0
−2� + � = 7 ⎩
⎪
⎨
⎪
⎧
� = 1
� = −2
� = 0
� − 2(0) = 0
−2� + � = 7 ⎩
⎪
⎨
⎪
⎧
� = 1
� = −2
� = 0
� = 0
−2(0) + � = 7 ⎩
⎪
⎨
⎪
⎧
� = 1
� = −2
� = 0
� = 0
� = 7
Se �(�) = ��� + ��� + �� + �, �� ��� :
�(�) = 1�³− 2�²+ 0� + 0 �(�) = �³− 2�²
�� �(�) = �, � ���� : �(�) = 7 RESPOSTA: �(�) = �³− 2�² � �(�) = 7
c) Usando o Método de Briot-Ruffini
a Coeficientes do dividendo
2 1 -4 4 7
1 2(1)-4 2(-2)+4 2(0)+7
1 -2 0 7
Coeficientes do quociente Resto
���� : �(�) = ��� + ��� + �� + �, � ���� : �(�) = �� − 2��
���� : �(�) = � �(�) = 7
15) Dado o polinômio �(�) = �³ – 2�² + �� – 1, onde � ∈ � � e seja
�(�)� ��� �� �� � ���� � = �. �� �(2) = 3.�(0), ��� ã� �(�) é ���� � �:
SOLUÇÃO:
�(�) = � �– 2� � + �� – 1 ∴ �(2) = 3.�(0)
(2)�– 2(2)� + �2 – 1 = 3[(0)�– 2(0) + �(0) – 1 ]
8 − 8 + 2� − 1 = −3
2� − 1 = −3 � = −1
16) O quociente da divisão de �(�) = 4�� – 4�³+ � – 1 ��� �(�) = 4�³ + 1 é:
SOLUÇÃO:
4�� – 4�³ + � – 1 4�³ + 1
−4�� − � � − 1
– 4�³ − 1 Resposta:
LETRA b)
Q(x) = � − 1
4�³ + 1
0
17) Qual o resto da divisão do polinômio �³ – 2�² + � + 1 ��� �² – � + 2 ?
SOLUÇÃO:
�³ – 2�² + � + 1 �² – � + 2
−�³+ �²− 2� � − 1
−�² − � + 1
R(x) = −2� + 3
�² − � + 2
−2� + 3
18) O quociente da divisão de �(�) = �³ – 7�² + 16� – 12 ��� �(�) = � – 3 é:
3 1 -7 16 -12
1 3(1)-7
-4
3(-4)+16
4
3(4)-12
0
�(�) = �� − 4� + 4
19) (UFU-MG) – Dividindo-se um polinômio f por (�– 3) , resulta um resto (– 7) e um
quociente (�– 4) . O polinômio é:
� ���ÇÃ�:
�(�) = (� − 3)(� − 4) − 7 = �� − 4� − 3� + 12 − 7 = �� − 7� + 5
20) Determine os valores de m, n e p, de modo que os polinômios��(�)� ��(�) sejam
idênticos, sendo: ��(�) = (� + � + �)�
� − (� + 1)�� + ��� + (� − �)� + � ;
��(�) = 2�
� + (2� + 7)� � + 5�� + 2�
(� + � + �)�� − (� + 1)�� + ��� + (� − �)� + � ≡ 2�� + (2� + 7)�� + 5�� + 2�
�
� + � + � = 0
1 + 2 − 3 = 0
��
�
−(� + 1) = 2
−� − 1 = 2
� = −3
�
� = 2� + 7
� = 2(−3) + 7
� = 1
�
5� = � − �
5(1) = � − (−3)
� = 2
�
� = 2�
� = 2(1)
� = 2
21) Determine A, B e C, sabendo que
5 − 3�
�� − 5� � + 6�
=
�
�
+
�
� − 2
+
�
� − 3
5 − 3�
�� − 5�� + 6�
=
�(� − 2)(� − 3) + �� (� − 3) + ��(� − 2)
�(� − 2)(� − 3)
5 − 3�
�� − 5�� + 6�
=
�� � − 5�� + 6� + ��� − 3�� + ��� − 2��
� � − 5� � + 6�
5 − 3� = (� + � + �)�� + (−5� − 3� − 2�)� + 6�
�
�
� + � + � = 0
5
6
+ � + � = 0
6� + 6� = −5(2)
12� + 12� = −10
�
−5� − 3� − 2� = −3
−5 �
5
6
� − 3� − 2� = −3
−18� − 12� = 7
��
6� = 5
� =
5
6 ⎩
⎪
⎨
⎪
⎧
12� + 12� = −10
−18� − 12� = 7
−6� = −3
� =
1
2
�
�
6� + 6� = −5
6 �
1
2
� + 6� = −5
� = −
4
3
22) Determine as soluções da equação, onde �(�) = 0, ���� �(�) é o quociente da
divisão do polinômio �(�) �� − 10�� + 24�� + 10� − 24 ��� �� − 6� + 5 .
�� − 10�� + 24�� + 10� − 24 �� − 6� + 5
−�� + 6�³ − 5��
−4�³+ 19�� + 10� − 24
−4�³− 24�² + 20�
−5�� + 30� − 24
−5�� − 30� + 25
1�������������
���� �
�(�) = �� − 4� − 5���������
����� ��� � (�)
�� − 4� − 5 = 0
� =
4 ± 6
2
�� = −1 ; ��� = 5
23) (UEPG – PR) – Os valores de � � � que tornam idênticos os polinômios
��(�) = �
�– �– 6 e ��(�) = (� + �)
�– � são, respectivamente?
�
��– �– 6 ≡ (� + �)�– �
��– �– 6 ≡ �� + 2�� + ��– �
��– �– 6 ≡ �� + 2�� + ��– �
��
2� = −1
� = −
1
2 �
�
�� − � = −6
�−
1
2
�
�
− � = −6
1 − 4� = −24
� =
25
4
24) (UFMG) – Determine o quociente e o resto da divisão de
�(�) = 4��– 4�� + �– 1 por �(�) = 4�� + 1.
4��– 4�� + �– 1 4�� + 1
−4�� − �
−4�³ − 1
4�³ + 1
0⏟
���� �
� − 1���
������ �� �
25) (CESCEM-SP) – Dividindo � �– 4�� + 7�– 3 por um certo polinômio �(�),
obtemos como quociente �– 1 e resto 2�– 1. O polinômio �(�) é igual a?
��– 4�� + 7�– 3 ≡ (��� + �� + �)(� − 1) + 2� − 1
��– 4�� + 7�– 3 ≡ ��� − ��� + ��� − �� + �� − � + 2� − 1
��– 4�� + 7�– 3 ≡ ��� + (−� + �)�� + (−� + � + 2)� + (−� − 1)
�
� = 1
� = 1
�
−� + � = −4
−1 + � = −4
� = −3
�
−� + � + 2 = 7
−(−3) + � + 2 = 7
� = 2
�
−� − 1 = −3
� = 2
�(�) = �� − 3� + 2
26) (UFPA) – O polinômio ��– 5� � + �� – � é divisível por ��– 3� + 6. Então, os
números m e n são tais que � + � é igual a?
��– 5� � + �� – � �� − 3� + 6
−�� + 3�� − 6�
−2� � + (� − 6)� − �
2�� − 6� + 12
(� − 12)� + (12 − �)���������������
���� � = 0
�⏟ − 2
����� ��� �
(� − 12)� + (12 − �) = 0� + 0
� − 12 = 0 ; 12 − � = 0
� = 12 ; � = 12
� + � = 24
27) (CEFET – PR) – Se �(�– 3)(�– 2) + �� (� − 3) + ��(�– 2) = 12. Determine os
valores de �, � � �:
�(�– 3)(�– 2) + �� (� − 3) + ��(�– 2) = 12
��� − 5�� + 6� + ��� − 3�� + ��� − 2�� = 12
(� + � + �)�� + (−5� − 3� − 2�)� + 6� = 0�� + 0� + 12
�
� + � + � = 0
2 + � + � = 0
� + � = −2
2� + 2� = −4
�
−5� − 3� − 2� = 0
−5(2) − 3� − 2� = 0
−3� − 2� = 10
�
6� = 12
� = 2
�
2� + 2� = −4
−3� − 2� = 10
−� = 6
� = −6
�
� + � = −2
−6 + � = −2
� = 4
28). (UNICAMP-SP) – Determine o quociente e o resto da divisão do polinômio
�(�) = ��– 2�� + 4 pelo polinômio �(�) = � �– 4.
��– 2�� + 4 ��– 4
−�� + 4�
−2�� + 4� + 4
2�� − 8
4� − 4���
���� �
�⏟ − 2
����� ��� �
29). (UFPR) – Se os polinômios �(�) = 4�� − (� + 2)�� − 5 e
�(�) = �� � + 5��– 5 são idênticos, então � �– � � é:
�(�) ≡ �(�)
4�� − (� + 2)�� − 5 ≡ �� � + 5��– 5
������ ����:
�|� = 4 �
−(� + 2) = 5
−� − 2 = 5
� = −7
�
� � − � �
(−7)� − (4)�
−343 − 64 = −407
30). (UFSE) – Dividindo-se o polinômio � = �� pelo polinômio � = ��– 1, obtém-se
quociente e resto, respectivamente, iguais a?
� � ��– 1
−�� + ��
��
−�� + 1
1⏟
���� �
�� + 1���
����� ��� �
31) (UFGO) – Se o polinômio � � + ���– 2� + 3 é divisível pelo polinômio
��– � + 1 , então o quociente é?
� � + ���– 2� + 3 � �– � + 1
−� � + � � − �
(� + 1) �� − 3� + 3
−(� + 1) � � + (� + 1) � − (� + 1)
(� − 2)� − (� − 2)
���� �
� + (� + 1)�������
������� � �
�
�
(� − 2)� − (� − 2) = 0� + 0
� − 2 = 0 �� − (� − 2) = 0
� = 2
�(�) = � + (� + 1)
�(�) = � + (2 + 1)
�(�) = � + 3
32) Dados os polinômios �(�) = �� + �� − � + 1 � �(�) = −3�� − � − 2,
calcule A �
1
2
� − B(−1).(01 ���� �)
A �
1
2
� = �
1
2
�
�
+ �
1
2
�
�
− �
1
2
� + 1 =
1
8
+
14
−
1
2
+ 1 =
1 + 2 − 4 + 8
8
=
7
8
B(−1) = −3(−1)� − (−1) − 2 = −3 + 1 − 2 = −4
A �
1
2
� − B(−1) =
7
8
− (−4) =
7
8
+ 4 =
39
8
33) Discuta o grau do polinômio �(�) = (� + 4)�� + (�� − 16)�� + (� − 4)� + 4.
�
���� 03:
� + 4 ≠ 0
� ≠ −4
�
�
���� 02
� + 4 = 0
� = −4
�� − 16 ≠ 0
� ≠ ±4
∄ � ∈ ℝ
�
�
���� 01
� = −4
� = ±4
� − 4 ≠ 0
� ≠ 4
� = −4
�
�
���� 0
� = −4
� = ±4
� = 4
∄ � ∈ ℝ
34) Determine o valor de � � � de modo que a divisão de do polinômio
�� − 9�� + 30�� − �� + � ��� (3 − �)(3 + �), seja exata.
[� − (3 − �)][� − (3 + �)] =
(� − 3 + �)(� − 3 − �) =
�� − 3� − � � − 3� + 9 + 3� + � � − 3� + 1 =
�� − 6� + 10
�� − 9�� + 30�� − �� + � �� − 6� + 10
−�� + 6�� − 10��
−3�� + 20�� − �� + �
3�� − 18�� + 30�
2�� + (30 − �)� + �
−2�� + 12� − 20
(42 − �)� + (� − 20)���������������
�� ��� (�)
�� − 3� + 2���������
����� ��� �(�)
��
�� �� �ã� ���� � → �(�) = 0
(42 − �)� + (� − 20) = 0� + 0
42 − � = 0 ∴ � = 42
� − 20 = 0 ∴ � = 20
35) Dois polinômios �(�) � �(�) têm graus n e m, respectivamente. Sabendo que o
grau de �(�) ∙ �(�) é 7, e que � − � = −1, determine o grau de �(�) + �(�).
�
�� �(�) ∙ �(�) é 7
������� �������� :
�� … ∙ � � … = �� …
� + � = 7
�
�
� + � = 7
−� + � = −1
2� = 6
� = 3
� = 4
��
�(�) = �� …
�(�) = �� …
�(�) + �(�)
�� … + �� …
�
� ��� �� ���� �� ∶
�(�) + �(�)
é
4
36) Um polinômio P(x) dividido por (� + 2) dá resto 4 e divido por (� − 1) dá resto
8. Determinar o resto da divisão de P(x) por (� + 2) ∙ (� − 1).
�(�) = (� + 2) ∙ (� − 1)�����������
�� �� ���
∙ �(�)�
����� � �� �
+ �� + ������
���� �
�(−2) = 4 → (−2 + 2) ∙ (−2 − 1) ∙ �(�) + �(−2) + � = 4 → −2� + � = 4
�(1) = 8 → (1 + 2) ∙ (1 − 1) ∙ �(�) + �(1) + � = 8 → � + � = 8
⎩
⎪
⎨
⎪
⎧ 2� − � = −4
� + � = 8
3� = 4
� =
4
3
�
�
2� − � = −4
2 �
4
3
� − � = −4
8 − 3� = −12
� =
20
2
��
�(�) = �� + �
�(�) =
4
3
� +
20
3
37) Determine o quociente e o resto da divisão de �(�) = 4�� + 2�� − � + 3 por
ℎ(�) = 3� − 2
a) Método da chave; (01 ���� �)
b) Método dos coeficientes a Determinar (Descartes); (01 ���� �)
c) Algoritmo de Briot-Ruffini. (01 ���� �)
a) Método da chave:
4�� + 2�� − � + 3 3� − 2
−4�� +
8
3
��
8
3
�� + 2�� − � + 3
−
8
3
�� +
16
9
��
34
9
�� − � + 3
−
34
9
� � +
68
27
�
41
27
� + 3
−
41
27
� +
82
81
325
81���������������������
���� � (�)
4
3
�� +
8
9
�� +
34
27
� +
41
81
�������������������
����� ��� �(�)
b) Método dos coeficientes a Determinar (Descartes);
�(�) ≡ ℎ(�) ���
���� ���
∙ �(�)�
����� ��� �
+ �(�) ���
���� �
4�� + 2�� − � + 3 ≡ (3� − 2)(��� + ��� + �� + �) + (�)
4�� + 2�� − � + 3 ≡ 3��� + 3��� + 3��� + 3�� − 2��� − 2��� − 2�� − 2� + �
4� � + 2�� − � + 3 ≡ 3�� � + (3� − 2�)� � + (3� − 2�)�� + (3� − 2�)� + (−2� + �)
�
�
3� = 4
� =
4
3
�
�
3� − 2� = 0
3� − 2 �
4
3
� = 0
� =
8
3
∙
1
3
� =
8
9
�
�
3� − 2� = 2
3� − 2 �
8
9
� = 2
� = �2 +
16
9
� ∙
1
3
� =
34
27
�
�
3� − 2� = −1
3� − 2 �
34
27
� = −1
� =
−1 +
68
27
3
=
41
27
3
� =
41
81
�
�
−2� + � = 3
−2 �
41
81
� + � = 3
� = 3 +
82
81
� =
325
81
����� �� ��(�) =
4
3
�� +
8
9
�� +
34
27
� +
41
81
� ��� ��(�) =
325
81
b) Algoritmo de Briot-Ruffini.
2
3
4 0 2 -1 3
4
2
3
∙ 4 + 0
2
3
∙
8
3
+ 2
2
3
∙
34
9
− 1
2
3
∙
41
27
+ 3
8
3
34
9
41
27
325
81
��(�) = 4�� +
8
3
�� +
34
9
� +
41
27
�(�) =
325
81
�(�) ≡ (3� − 2) ∙ �(�) + �(�) ≡ �� − �
2
3
�� ∙ 3 ∙ �(�)�����
��(�)
+ �(�)
3 ∙ �(�) = �′(�) → �(�) =
�′(�)
3
�(�) =
��(�)
3
→ �(�) =
4�� +
8
3
�� +
34
9
� +
41
27
3
�(�) =
4
3
�� +
8
9
�� +
34
27
� +
41
81
38) Determine o quociente e o resto da divisão de �(�) = 4� � + 2� � − � + 3 por
H(�) = 3� − 2
a) Método da chave; (01 ���� �)
b) Método dos coeficientes a Determinar (Descartes); (01 ���� �)
4� � + 2� � − � + 3 3� − 2
−4� � +
8
3
��
8
3
�� + 2� � − � + 3
−
8
3
� � +
16
9
��
34
9
� � − � + 3
−
34
9
� � +
68
27
�
41
27
� + 3
−
41
27
� +
82
81
325
27�������������������
����� (�)
4
3
� � +
8
9
� � +
34
27
� +
41
81�����������������
�� ����� �� (�)
b) Método dos coeficientes a Determinar (Descartes);
�(�) ≡ ℎ(�) ���
�� �� ���
∙ �(�)�
�� ����� ��
+ �(�) ���
�����
4� � + 2�� − � + 3 ≡ (3� − 2 )(�� � + �� � + �� + �) + (�)
4� � + 2�� − � + 3 ≡ 3�� � + 3�� � + 3�� � + 3�� − 2�� � − 2�� � − 2�� − 2� + �
4� � + 2�� − � + 3 ≡ 3�� � + (3� − 2�)� � + (3� − 2�)�� + (3� − 2�)� + (−2� + �)
�
�
3� = 4
� =
4
3
�
�
3� − 2� = 0
3� − 2 �
4
3
� = 0
9� − 8 = 0
� =
8
9
�
�
3� − 2� = 2
3� − 2 �
8
9
� = 2
27� − 16 = 18
� =
34
27
�
�
3� − 2� = −1
3� − 2 �
34
27
� = −1
81� − 68 = −27
� =
41
81
�
�
−2� + � = 3
−2 �
41
81
� + � = 3
−82 + 81� = 243
� =
325
81
�(�) =
4
3
� � +
8
9
� � +
34
27
� +
41
81
; �(�) =
325
81
39) Discuta o grau do polinômio (�� − 2� − 3)�� − (�� − 9)� − (� + 2)
�
���� 3
�� − 2� − 3 ≠ 0
� ≠ −1 �� � ≠ 3
� = {� ∈ ℝ | � ≠ −1 �� � ≠ 3}
�
���� 2
∄ � �� ��
�� ����
2
�
�
���� 1
� = −1 �� � = 3
�� − 9 ≠ 0
� ≠ −3 �� � ≠ 3
� = −1
�
���� 0
� = 3
40) Sejam os polinômios �(�) = (3� + 2)� + 2 � �(�) = 2�� − 3� + 1 nos quais a
é uma constante. Determine o valor de a para que o polinômio � ∙ � tenha grau 2.
� ∙ � = [(3� + 2)� + 2] ∙ [2�� − 3� + 1]
� ∙ � = (6�� + 4�)�� + (−9�� − 6�)� + (3� + 2)� + 4�� − 6� + 2
� ∙ � = (6�� + 4�)�� − (9�� − � − 2)� − 6� + 2
�� �� 2 → 6�� + 4� ≠ 0 → �(6� + 4) ≠ 0 → � ≠ 0 � � ≠ −
2
3
41) Sabendo que os polinômios �(�) = �� + 3� �+ � � �(�) = �� + � �− 4 são
divisíveis por �(�)� + 1, Calcule o valor de � + �.
�4 + 3� �+ � � + 1
−� � − � �
−�� + 3�� + �
� � + � �
� � + 3�� + �
−�� − �
(3� − 1)� + �
−(3� − 1)� − (3� − 1)
(� − 3� + 1)�����������������
����� (�)
�� − �� + � + (3� + 1)�����������������
�� ����� �� (�)
�2 + � �− 4 � + 1
−� � − �
(� − 1)� − 4
−(� − 1)� − (� − 1)
−� − 3�����
����� (�)
� + (� − 1)���������
�� ����� �� (�)
−� − 3 = 0 → � = −3
� − 3� + 1 = 0
� − 3(−3)+ 1 = 0
� = −10
� + � = −3 − 10 = −13
42) O polinômio �(�) = �� , � � � � ∈ ℝ, corresponde ao resto da divisão de
�(�) = 4�� − (� − 5)� � + 4� − 2 por ℎ(�) = � � − 2� + �. Determine os valores de
� � � e escreva o polinômio �(�).
4�3 − (� − 5)�2 + 4� − 2 �2 − 2� + �
−4� � + 8� � − 4� �
−(� − 13)�� + (4 − 4�)� − 2
+(� − 13)�� − 2(� − 13)� + (� − 13)�
(−2� + 30 − 4�)� − 2 + (� − 13)������������������������
����� (�)
4� − (� − 13)���������
�� ����� �� (�)
�
�
�(�) = ��
(−2� + 30 − 4�)� − 2 + (� − 13)� = 0 + ��
−2� + 30 − 4� = 0 −2 + (� − 13)� = ��
� = 15 − 2� −2 + [(15 − 2�) − 13]� = ��
−2 + (2 − 2�)� = ��
−2 + 2� + 2 = ��
� = 2
�
�(�) = ��
�(�) = 2�