#### Answer

$6(p+t)(p-t)$

#### Work Step by Step

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}