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

$3(2x+5)(x+1)$
Factoring the $GCF= 3 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 6x^2+21x+15 \\\\= 3(2x^2+7x+5) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 3(2x^2+7x+5) \end{array} has $ac= 2(5)=10$ and $b= 7 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 5,2 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 3(2x^2+5x+2x+5) .\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} 3[(2x^2+5x)+(2x+5)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 3[x(2x+5)+(2x+5)] .\end{array} Factoring the $GCF= (x+6)$ of the entire expression above results to \begin{array}{l}\require{cancel} 3[(2x+5)(x+1)] \\\\= 3(2x+5)(x+1) .\end{array}