Answer
$(3c+20)(3c+4)$
Work Step by Step
Let $z=(3c+6)$. Then the given expression, $
(3c+6)^2+12(3c+6)-28
$, is equivalent to $
z^2+12z-28
$.
The two numbers whose product is $ac=
1(-28)=-28
$ and whose sum is $b=
12
$ are $\{
14,-2
\}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $
z^2+12z-28
$, is
\begin{array}{l}\require{cancel}
z^2+14z-2z-28
\\\\=
(z^2+14z)-(2z+28)
\\\\=
z(z+14)-2(z+14)
\\\\=
(z+14)(z-2)
.\end{array}
Since $z=(3c+6)$, then,
\begin{array}{l}
(3c+6+14)(3c+6-2)
\\\\=
(3c+20)(3c+4)
.\end{array}
