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

$2(s+t)(5s-3t)$
Factoring the $GCF= 2 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 10s^2+4st-6t^2 \\\\= 2(5s^2+2st-3t^2) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 2(5s^2+2st-3t^2) \end{array} has $ac= 5(-3)=-15$ and $b= 2 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 5,-3 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 2(5s^2+5st-3st-3t^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} 2[(5s^2+5st)-(3st+3t^2)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 2[5s(s+t)-3t(s+t)] .\end{array} Factoring the $GCF= (s+t)$ of the entire expression above results to \begin{array}{l}\require{cancel} 2[(s+t)(5s-3t)] \\\\= 2(s+t)(5s-3t) .\end{array}