## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 9 - Cumulative Review Exercises - Page 1038: 14

#### Answer

a. $\pm1,\pm3,\pm\frac{1}{2},\pm\frac{3}{2},\pm\frac{1}{4},\pm\frac{3}{4},\pm\frac{1}{8},\pm\frac{3}{8},\pm\frac{1}{16},\pm\frac{3}{16},\pm\frac{1}{32},\pm\frac{3}{32}$ b. $-\frac{1}{8}, \frac{3}{4}, 1$ #### Work Step by Step

a. Based on the given equation, we have $p=\pm1,\pm3$ and $q=\pm1,\pm2,\pm4,\pm8,\pm16,\pm32$ Thus the possible rational zeros are $\frac{p}{q}=\pm1,\pm3,\pm\frac{1}{2},\pm\frac{3}{2},\pm\frac{1}{4},\pm\frac{3}{4},\pm\frac{1}{8},\pm\frac{3}{8},\pm\frac{1}{16},\pm\frac{3}{16},\pm\frac{1}{32},\pm\frac{3}{32}$ b. Using the given information, we can identify $x=1$ as a zero. Now, we use synthetic division as shown. We can factor the equation as $(x-1)(32x^2-20x-3)=0$ or $(x-1)(8x+1)(4x-3)=0$ Thus the solutions are $x=-\frac{1}{8}, \frac{3}{4}, 1$

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