## Intermediate Algebra: Connecting Concepts through Application

$(3p+4)(2p+5)$
$\bf{\text{Solution Outline:}}$ To factor the given expression, $6p^2+23p+20 ,$ 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. $\bf{\text{Solution Details:}}$ Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $6(20)=120$ and the value of $b$ is $23 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ 8,15 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 6p^2+8p+15p+20 .\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} (6p^2+8p)+(15p+20) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 2p(3p+4)+5(3p+4) .\end{array} Factoring the $GCF= (3p+4)$ of the entire expression above results to \begin{array}{l}\require{cancel} (3p+4)(2p+5) .\end{array}