Answer
$(p-6q)(3p+2q)$
Work Step by Step
Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{
expression
}$
\begin{array}{l}\require{cancel}
3p^2-16pq-12q^2
\end{array} has $ac=
3(-12)=-36
$ and $b=
-16
.$
The two numbers with a product of $c$ and a sum of $b$ are $\left\{
-18,2
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
3p^2-18pq+2pq-12q^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}
(3p^2-18pq)+(2pq-12q^2)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
3p(p-6q)+2q(p-6q)
.\end{array}
Factoring the $GCF=
(p-6q)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(p-6q)(3p+2q)
.\end{array}
