## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

# Chapter R - Elementary Algebra Review - R.5 Polynomials and Factoring - R.5 Exercise Set - Page 970: 23

#### Answer

$(n-2)(m+3)$

#### Work Step by Step

Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} mn-2m+3n-6 \\\\= (mn-2m)+(3n-6) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} (mn-2m)+(3n-6) \\\\= m(n-2)+3(n-2) .\end{array} Factoring the $GCF= (n-2)$ of the entire expression above results to \begin{array}{l}\require{cancel} (n-2)(m+3) .\end{array}

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