Explanation Procedure used: Procedure for Synthetic Division: “1. Write the coefficients of f ( x ) . Be certain that the powers are in descending order and that zeros are inserted for missing powers. 2. Carry down the left coefficient, then multiply it by r , and place this product under the second coefficient of the top line. 3. Add the two numbers in the second column and place the result below. Multiply this sum by r and place the product under the third coefficient of the top line. 4. Continue this process until the bottom row has as many numbers as the top row”. The Factor Theorem: “If f ( r ) = 0 , then x = r is a zero of f ( x ) , x − r is a factor of f ( x ) and x = r is a root of the equation f ( x ) = 0 ”. Calculation: Divide the polynomial 2 Z 4 − Z 3 − 4 Z 2 + 1 by 2 Z − 1 as follows. The coefficient of Z in the possible factor is not 1, so the factor is not of the form Z − r . The divisor can be written as 2 Z − 1 = 2 ( Z − 1 2 ) . If 2 ( Z − 1 2 ) is a factor of the function, 2 Z − 1 is a factor. That is, if Z = 1 2 , the remainder is zero, then Z − 1 2 is a factor of 2 Z 4 − Z 3 − 4 Z 2 + 1 . The powers of Z are in descending order, write down the coefficient of f ( Z ) , write 1 2 for the divisor Z − 1 2 and place 1 2 to the right. 2 − 1 − 4 0 1 1 2 Carry the left coefficient 2 to the bottom line and multiply it by 1 2 , and place the product 1 under the second coefficient − 1 of the top line and proceed in this similar manner to find the remainder. 2 − 1 − 4 0 1 1 2 1

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Chapter 15.1, Problem 37E

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Whether 2Z−1 is a factor of 2Z4−Z3−4Z2+1 or not by using factor theorem and synthetic division.

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Chapter 15.1, Problem 36E

Chapter 15.1, Problem 38E

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Chapter 15 Solutions

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Whether 2 Z − 1 is a factor of 2 Z 4 − Z 3 − 4 Z 2 + 1 or not by using factor theorem and synthetic division.

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Whether 2Z−1 is a factor of 2Z4−Z3−4Z2+1 or not by using factor theorem and synthetic division.

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Chapter 15 Solutions