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: 47


$x=2$, $x=-3,-1,2$

Work Step by Step

Step 1. With the given graph, we can determine a solution of the equation as $x=2$. Step 2. The coefficients of the dividend $f(x)=x^3+2x^2-5x-6$ can be identified in order as $\{1,2,-5,-6\}$ and the divisor is $x-2$; use synthetic division as shown in the figure to get the quotient and the remainder. Step 3. We can identify the result as $f(x)=(x^2+4x+3)(x-2)=(x+1)(x+3)(x-2)$; thus we can find the zeros of $f(x)$ as $x=-3,-1,2$
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