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: 53

Answer

$6a(a-3)(a+5)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ First, factor the GCF of the terms. Then, to factor the quadratic expression $x^2+bx+c,$ find two numbers, $m_1$ and $m_2,$ whose product is $c$ and whose sum is $b$. Express the factored form as $GCF(x+m_1)(x+m_2).$ $\bf{\text{Solution Details:}}$ Factoring the $GCF= 6a ,$ the given expression, $ 6a^3+12a^2-90a ,$ is equivalent to \begin{array}{l}\require{cancel} 6a(a^2+2a-15) .\end{array} In the expression above, the value of $c$ is $ -15 $ and the value of $b$ is $ 2 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,-15\}, \{3,-5\}, \{-1,15\}, \{-3,5\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ -3,5 \}.$ Hence, the factored form of the given expression is \begin{array}{l}\require{cancel} 6a(a-3)(a+5) .\end{array}
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