Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 3 - Polynomial and Rational Functions - Section 3.2 The Real Zeros of a Polynomial Function - 3.2 Assess Your Understanding - Page 224: 39

Answer

$\pm \dfrac{1}{2}, \pm \dfrac{1}{3}, \pm \dfrac{9}{2}, \pm \dfrac{1}{6}, \pm \dfrac{3}{2}, \pm 1, \pm 3, \pm 9$

Work Step by Step

Let us consider that $m$ is a factor of the constant term and $n$ is a factor of the leading coefficient. Then the potential zeros can be expressed by the possible combinations as: $\dfrac{m}{n}$. We see from the given polynomial function that it has a constant term of $9$ and a leading coefficient of $6$. The possible factors $m$ of the constant term and $n$ of the leading coefficient are: $m=\pm 1, \pm 3, \pm 9$ and $n=\pm 1, \pm 2, \pm 3, \pm 6$, Therefore, the possible rational roots of $f(x)$ are: $\dfrac{m}{n}=\pm \dfrac{1}{2}, \pm \dfrac{1}{3}, \pm \dfrac{9}{2}, \pm \dfrac{1}{6}, \pm \dfrac{3}{2}, \pm 1, \pm 3, \pm 9$
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