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: 13

Answer

a. $\frac{p}{q}=\pm1,\pm2,\pm3,\pm6$ b. $x=-1$; see figure. c. $x=-1,\frac{-3\pm\sqrt {33}}{2}$

Work Step by Step

a. Given the function $f(x)=x^3+4x^2-3x-6$, we have $p=\pm1,\pm2,\pm3,\pm6$ and $q=\pm1$; thus the possible rational zeros are $\frac{p}{q}=\pm1,\pm2,\pm3,\pm6$ b. Starting from the easiest number, test the possible rational zeros using synthetic division; we can find $x=-1$ as a zero shown in the figure. c. Based on the results from part-b, we have $f(x)=x^3+4x^2-3x-6=(x+1)(x^2+3x-6)$ Thus the zeros are $x=-1,\frac{-3\pm\sqrt {33}}{2}$
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