Intermediate Algebra: Connecting Concepts through Application

$(8p-3)(3p-4)$
$\bf{\text{Solution Outline:}}$ To factor the given expression, $24p^2-41p+12 ,$ 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 $24(12)=288$ and the value of $b$ is $-41 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -9,-32 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 24p^2-9p-32p+12 .\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} (24p^2-9p)-(32p-12) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 3p(8p-3)-4(8p-3) .\end{array} Factoring the $GCF= (8p-3)$ of the entire expression above results to \begin{array}{l}\require{cancel} (8p-3)(3p-4) .\end{array}