## Intermediate Algebra: Connecting Concepts through Application

$4m(2n+3)(5n-7)$
$\bf{\text{Solution Outline:}}$ To factor the given expression, $40mn^2+4mn-84m ,$ factor first the $GCF.$ Then 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. $\bf{\text{Solution Details:}}$ Factoring the $GCF= 4m ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 4m(10n^2+n-21) .\end{array} Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $10(-21)=-210$ and the value of $b$ is $1 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ 15,-14 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 4m(10n^2+15n-14n-21) .\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} 4m[(10n^2+15n)-(14n+21)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 4m[5n(2n+3)-7(2n+3)] .\end{array} Factoring the $GCF= (2n+3)d$ of the entire expression above results to \begin{array}{l}\require{cancel} 4m[(2n+3)(5n-7)] \\\\= 4m(2n+3)(5n-7) .\end{array}