Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 5 - Section 5.2 - Factoring Trinomials - 5.2 Exercises: 13

Answer

$(x-2a)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ x^2+ax-6a^2 ,$ find two numbers, $m_1$ and $m_2,$ whose product is $c$ and whose sum is $b$ in the quadratic expression $x^2+bx+c.$ Then, express the factored form as $(x+m_1)(x+m_2).$ $\bf{\text{Solution Details:}}$ In the given expression, the value of $c$ is $ -6 $ and the value of $b$ is $ 1 .$ The possible pairs of integers whose product is $c$ are \begin{array}{l}\require{cancel} \{ 1,-6 \}, \{ 2,-3 \}, \{ -1,6 \}, \{ -2,3 \} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ -2,3 \}.$ Hence, the factored form of the given expression is \begin{array}{l}\require{cancel} (x-2a)(x+3a) .\end{array} The missing factor in the given expression is $ (x-2a) .$
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