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: 31



Work Step by Step

$\bf{\text{Solution Outline:}}$ 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. $\bf{\text{Solution Details:}}$ In the given expression, $ -6m^2-13m+15 ,$ the value of $ac$ is $ -6(15)=-90 $ and the value of $b$ is $ -13 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,-90\}, \{2,-45\}, \{3,-30\}, \{5,-18\}, \{6,-15\}, \{9,-10\}, \{-1,90\}, \{-2,45\}, \{-3,30\}, \{-5,18\}, \{-6,15\}, \{-9,10\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ 5,-18 \}.$ Using these $2$ numbers to decompose the middle term of the given expression results to \begin{array}{l}\require{cancel} -6m^2+5m-18m+15 .\end{array} Grouping the first and third terms and the second and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (-6m^2+5m)-(18m-15) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} -m(6m-5)-3(6m-5) .\end{array} Factoring the $GCF= (6m-5) $ of the entire expression above results to \begin{array}{l}\require{cancel} (6m-5)(-m-3) .\end{array}
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