Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Section 2.4 - Dividing Polynomials; Remainder and Factor Theorems - Exercise Set - Page 364: 46


$x=-4, -\frac{1}{3}, 2$

Work Step by Step

Step 1. As $-\frac{1}{3}$ is a zero of $f(x)$, we can try to divide $f(x)$ by $x+\frac{1}{3}$ Step 2. The coefficients of the dividend $f(x)$ can be identified in order as $\{3,7,-22,-8\}$ and the divisor is $x+\frac{1}{3}$; use synthetic division as shown in the figure to get the quotient and the remainder. Step 3. We can identify the result as $f(x)=(3x^2+6x-24)(x+\frac{1}{3})=3(x^2+2x-8)(x+\frac{1}{3})=3(x+4)(x-2)(x+\frac{1}{3})$; thus we can find the zeros of $f(x)$ as $x=-4, -\frac{1}{3}, 2$
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