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

$t(t+2)(7t+1)$
Factoring the $GCF= t ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 7t^3+15t^2+2t \\\\= t(7t^2+15t+2) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} t(7t^2+15t+2) \end{array} has $ac= 7(2)=14$ and $b= 15 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 14,1 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} t(7t^2+14t+1t+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} t[(7t^2+14t)+(1t+2)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} t[7t(t+2)+(t+2)] .\end{array} Factoring the $GCF= (t+2)$ of the entire expression above results to \begin{array}{l}\require{cancel} t[(t+2)(7t+1)] \\\\= t(t+2)(7t+1) .\end{array}