Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 4 - Polynomial and Rational Functions - Chapter Review - Review Exercises - Page 244: 33

Answer

$-2,\frac{1}{2},\frac{1}{2}$ $f(x)=(x+2)(2x-1)^2$

Work Step by Step

Step 1. Based on the Rational Zeros Theorem, list possible rational zeros $\frac{p}{q}=\pm1,\pm2,\pm\frac{1}{2},\pm\frac{1}{4}$ Step 2. Use synthetic division to find one or more zeros $x=-2$ as shown in the figure. Step 3. Use to the quotient to solve $4x^2-4x+1=0$ or $(2x-1)^2=0$, thus $x=\frac{1}{2},\frac{1}{2}$ Step 4. We can factor the function as $f(x)=(x+2)(2x-1)^2$
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