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

$5(a+3)(2a-1)$
Factoring the $GCF= 5 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 10a^2+25a-15 \\\\= 5(2a^2+5a-3) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 5(2a^2+5a-3) \end{array} has $ac= 2(-3)=-6$ and $b= 5 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 6,-1 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 5(2a^2+6a-1a-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} 5[(2a^2+6a)-(a+3)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 5[2a(a+3)-(a+3)] .\end{array} Factoring the $GCF= (a+3)$ of the entire expression above results to \begin{array}{l}\require{cancel} 5[(a+3)(2a-1)] \\\\= 5(a+3)(2a-1) .\end{array}