Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 3 - Polynomial and Rational Functions - 3.3 Zeros of Polynomial Functions - 3.3 Exercises - Page 336: 42

Answer

(a) $\pm1,\pm2,\pm4,\pm8$ (b) $\{-2,-1,4\}$ (c) $ƒ(x)=(x-4)(x+1)(x+2)$

Work Step by Step

(a) Based on the given polynomial, we have factors $p=\pm1,\pm2,\pm4,\pm8, q=\pm1$. We can list all possible rational zeros as: $\frac{p}{q}=\pm1,\pm2,\pm4,\pm8$ (b) Use synthetic division and the above possible values, find one zero $x=-1$ as shown in the figure. With quotient $x^2-2x-8=0$, we have $(x+2)(x-4)=0$ which gives $x=-2, 4$. Thus the zeros are $\{-2,-1,4\}$ (c) Based on the know zeros and the factor theorem, we can factor $ƒ(x)$ into linear factors as $ƒ(x)=(x-4)(x+1)(x+2)$
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