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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+ �� + � �� (�) < �� (�), ��� ã��� (�) < 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+ �� + � �� (�) < �� (�), ��� ã��� (�) < 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+ �� + � �� (�) < �� (�), ��� ã��� (�) < 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�
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