Precalculus (6th Edition) Blitzer

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

Chapter 2 - Section 2.5 - Zeros of Polynomial Functions - Exercise Set - Page 377: 15


a. $\frac{p}{q}=\pm1,\pm2,\pm\frac{1}{2}$ b. $x=-2$; see figure. c. $x=-2,\frac{-1\pm i}{2}$

Work Step by Step

a. Given the function $f(x)=2x^3+6x^2+5x+2$, we have $p=\pm1,\pm2$ and $q=\pm1,\pm2$ Thus the possible rational zeros are $\frac{p}{q}=\pm1,\pm2,\pm\frac{1}{2}$ b. Starting from the easiest number, test the possible rational zeros using synthetic division. We can find $x=-2$ as a zero (shown in the figure). c. Based on the results from part-b, we have $f(x)=2x^3+6x^2+5x+2=(x+2)(2x^2+2x+1)$ Thus the zeros are $x=-2,\frac{-1\pm i}{2}$
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