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: 46



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 $GCF= 2 ,$ the given expression, $ 36x^2+18x-4 ,$ is equivalent to \begin{array}{l}\require{cancel} 2(18x^2+9x-2) .\end{array} In the expression above, the value of $ac$ is $ 18(-2)=-36 $ and the value of $b$ is $ 9 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,-36\}, \{2,-18\}, \{3,-12\}, \{4,-9\}, \{6,-6\}, \{-1,36\}, \{-2,18\}, \{-3,12\}, \{-4,9\}, \{-6,6\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ -3,12 \}.$ Using these $2$ numbers to decompose the middle term of the expression above results to \begin{array}{l}\require{cancel} 2(18x^2-3x+12x-2) .\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[(18x^2-3x)+(12x-2)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 2[3x(6x-1)+2(6x-1)] .\end{array} Factoring the $GCF= (6x-1) $ of the entire expression above results to \begin{array}{l}\require{cancel} 2[(6x-1)(3x+2)] \\\\= 2(6x-1)(3x+2) .\end{array}
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