Answer
$3(y-7)(y+4)$
Work Step by Step
Factoring the $GCF=
3
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
3y^2-9y-84
\\\\=
3(y^2-3y-28)
.\end{array}
Using the factoring of trinomials in the form $x^2+bx+c,$ the $\text{
expression
}$
\begin{array}{l}\require{cancel}
3(y^2-3y-28)
\end{array} has $c=
-28
$ and $b=
-3
.$
The two numbers with a product of $c$ and a sum of $b$ are $\left\{
-7,4
\right\}.$ Using these two numbers, the $\text{
expression
}$ above is equivalent to
\begin{array}{l}\require{cancel}
3(y-7)(y+4)
.\end{array}