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 - Page 337: 47



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 $GCF= -5 ,$ the given expression, $ -15a^2-70a+120 ,$ is equivalent to \begin{array}{l}\require{cancel} -5(3a^2+14a-24) .\end{array} In the expression above, the value of $c$ is $ 3(-24)=-72 $ and the value of $b$ is $ 14 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,-72\}, \{2,-36\}, \{3,-24\}, \{4,-18\}, \{8,-9\}, \{-1,72\}, \{-2,36\}, \{-3,24\}, \{-4,18\}, \{-8,9\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ -4,18 \}.$ Using these $2$ numbers to decompose the middle term of the given expression results to \begin{array}{l}\require{cancel} -5(3a^2-4a+18a-24) .\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} -5[(3a^2-4a)+(18a-24)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} -5[a(3a-4)+6(3a-4)] .\end{array} Factoring the $GCF= (3a-4) $ of the entire expression above results to \begin{array}{l}\require{cancel} -5[(3a-4)(a+6)] \\\\= -5(3a-4)(a+6) .\end{array}
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