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

$5(pt+6)(pt-1)$
Factoring the $GCF= 5$, then the given expression, $5p^2t^2+25pt-30$ is equivalent to \begin{array}{l} 5(p^2t^2+5pt-6) .\end{array} The two numbers whose product is $ac= 1(-6)=-6$ and whose sum is $b= 5$ are $\{ 6,-1 \}$. Using these two numbers to decompose the middle term of the expression, $5(p^2t^2+5pt-6) ,$ then the factored form is \begin{array}{l} 5(p^2t^2+6pt-1pt-6) \\\\= 5[(p^2t^2+6pt)-(1pt+6)] \\\\= 5[pt(pt+6)-(pt+6)] \\\\= 5[(pt+6)(pt-1)] \\\\= 5(pt+6)(pt-1) .\end{array}