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: 48



Work Step by Step

$\bf{\text{Solution Outline:}}$ First, factor the GCF of the terms. Then, to factor the quadratic expression $ax^2+bx+c,$ find two numbers whose product is $ac$ and whose sum is $b$. Use these $2$ numbers to decompose the middle term of the quadratic expression and then use factoring by grouping. $\bf{\text{Solution Details:}}$ Using the negative $GCF= -2 ,$ the given expression, $ -12a^2-10a+42 ,$ is equivalent to \begin{array}{l}\require{cancel} -2(6a^2+5a-21) .\end{array} In the expression above, the value of $c$ is $ 6(-21)=-126 $ and the value of $b$ is $ 5 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,-126\}, \{2,-63\}, \{3,-42\}, \{6,-21\}, \{9,-14\}, \{-1,126\}, \{-2,63\}, \{-3,42\}, \{-6,21\}, \{-9,14\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ -9,14 \}.$ Using these $2$ numbers to decompose the middle term of the given expression results to \begin{array}{l}\require{cancel} -2(6a^2-9a+14a-21) .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} -2[(6a^2-9a)+(14a-21)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} -2[3a(2a-3)+7(2a-3)] .\end{array} Factoring the $GCF= (2a-3) $ of the entire expression above results to \begin{array}{l}\require{cancel} -2[(2a-3)(3a+7)] \\\\= -2(2a-3)(3a+7) .\end{array}
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