## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 2 - Section 2.5 - Zeros of Polynomial Functions - Exercise Set - Page 377: 18

#### Answer

a. $\frac{p}{q}=\pm1,\pm2,\pm4$ b. $x=-1$ (see figure). c. $x=-1,4$

#### Work Step by Step

a. Given the function $f(x)=x^3-2x^2-7x-4$, we have $p=\pm1,\pm2,\pm4$ and $q=\pm1$ Thus the possible rational zeros are $\frac{p}{q}=\pm1,\pm2,\pm4$ 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 $f(x)=x^3-2x^2-7x-4=(x+1)(x^2-3x-4)=(x+1)(x-4)(x+1)$ Thus the zeros are $x=-1,4$

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