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

$6(p+t)(p-t)$
Factoring the $GCF= 6 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 6p^2-6t^2 \\\\= 6(p^2-t^2) .\end{array} The expressions $p^2$ and $t^2$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $p^2-t^2 ,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{array}{l}\require{cancel} 6[(p)^2-(t)^2] \\\\= 6[(p+t)(p-t)] \\\\= 6(p+t)(p-t) .\end{array}