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

$5(3a-4b)(a+b)$
Factoring the $GCF= 5 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 15a^2-5ab-20b^2 \\\\= 5(3a^2-ab-4b^2) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 5(3a^2-ab-4b^2) \end{array} has $ac= 3(-4)=-12$ and $b= -1 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ -4,3 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 5(3a^2-4ab+3ab-4b^2) .\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} 5[(3a^2-4ab)+(3ab-4b^2)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 5[a(3a-4b)+b(3a-4b)] .\end{array} Factoring the $GCF= (3a-4b)$ of the entire expression above results to \begin{array}{l}\require{cancel} 5[(3a-4b)(a+b)] \\\\= 5(3a-4b)(a+b) .\end{array}