## College Algebra (11th Edition)

Published by Pearson

# Chapter R - Review Exercises - Page 75: 68

#### Answer

$(3m+1)(2m-5)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $6m^2-13m-5 ,$ find two numbers whose product is $ac$ and whose sum is $b$ in the quadratic expression $ax^2+bx+c.$ Use these $2$ numbers to decompose the middle term of the given quadratic expression and then use factoring by grouping. $\bf{\text{Solution Details:}}$ In the trinomial expression above the value of $ac$ is $6(-5)=-30$ and the value of $b$ is $-13 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\{ 2,-15 \}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 6m^2+2m-15m-5 .\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} (6m^2+2m)-(15m+5) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 2m(3m+1)-5(3m+1) .\end{array} Factoring the $GCF= (3m+1)$ of the entire expression above results to \begin{array}{l}\require{cancel} (3m+1)(2m-5) .\end{array}

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