Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Section 2.4 - Dividing Polynomials; Remainder and Factor Theorems - Exercise Set - Page 364: 39



Work Step by Step

Step 1. Based on the Remainder Theorem, the value of $f(c)$ can be obtained as the remainder of $f(x)$ divided by $x-c$. Step 2. The coefficients of the dividend $f(x)$ can be identified in order as $\{2,-5,-1,3,2\}$ and the divisor is $x+\frac{1}{2}$; use synthetic division as shown in the figure to get the quotient and the remainder. Step 3. We can identify the result as $f(x)=(2x^3-6x^2+2x+2)(x+\frac{1}{2})+1$; thus we have $f(-\frac{1}{2})=1$.
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