## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 5 - Section 5.2 - Factoring Trinomials - 5.2 Exercises - Page 337: 39

#### Answer

$(6m-5)^2$

#### 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, $36m^2-60m+25 ,$ the value of $ac$ is $36(25)=900$ and the value of $b$ is $-60 .$ The $2$ numbers that have a product $ac$ and a sum of $b$ are $\{ -30,-30 \}.$ Using these $2$ numbers to decompose the middle term of the given expression results to \begin{array}{l}\require{cancel} 36m^2-30m-30m+25 .\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} (36m^2-30m)-(30m-25) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 6m(6m-5)-5(6m-5) .\end{array} Factoring the $GCF= (6m-5)$ of the entire expression above results to \begin{array}{l}\require{cancel} (6m-5)(6m-5) \\\\= (6m-5)^2 .\end{array}

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