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

$2y(4x-3)(3x+1)$
Factoring the $GCF=2y,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 24x^2y-6y-10xy \\\\= 2y(12x^2-3-5x) \\\\= 2y(12x^2-5x-3) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 2y(12x^2-5x-3) \end{array} has $ac= 12(-3)=-36$ and $b= -5 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ -9,4 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 2y(12x^2-9x+4x-3) .\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} 2y[(12x^2-9x)+(4x-3)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 2y[3x(4x-3)+(4x-3)] .\end{array} Factoring the $GCF= (4x-3)$ of the entire expression above results to \begin{array}{l}\require{cancel} 2y[(4x-3)(3x+1)] \\\\= 2y(4x-3)(3x+1) .\end{array}