## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 3 - Section 3.4 - Real Zeros of Polynomials - 3.4 Exercises - Page 283: 8

#### Answer

$-1, 1, -2, 2, -3, 3, -4, 4, -6, 6, -12, 12, -\frac{1}{2}, \frac{1}{2}, -\frac{1}{3}, \frac{1}{3}, -\frac{1}{6}, \frac{1}{6}, -\frac{2}{3}, \frac{2}{3}, -\frac{3}{2}. \frac{3}{2}, -\frac{4}{3}, \frac{4}{3}$

#### Work Step by Step

Note that: The factors of $12$ are $\pm1, \pm2, \pm3, \pm4, \pm6, \pm12$. The factors of $6$ are $\pm1, \pm2, \pm3, \pm6$ Thus, The possible values of $p$ are $\pm1, \pm2, \pm3, \pm4, \pm6, \pm12$. The possible values of $q$ are $\pm1, \pm2, \pm3, \pm6$ Therefore the possible rational zeros $\frac{p}{q}$ of $R(x)$ are: $\\-1, 1, -2, 2, -3, 3, -4, 4, -6, 6, -12, 12, -\frac{1}{2}, \frac{1}{2}, -\frac{1}{3}, \frac{1}{3}, -\frac{2}{3}, \frac{2}{3}, -\frac{3}{2}. \frac{3}{2}, -\frac{4}{3}, \frac{4}{3}-\frac{1}{6}, \frac{1}{6}$

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