Answer
$-4(x+3)(3x-2)$
Work Step by Step
Factoring the $GCF=
-4
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
-12x^2-28x+24
\\\\=
-4(3x^2+7x-6)
.\end{array}
Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{
expression
}$
\begin{array}{l}\require{cancel}
-4(3x^2+7x-6)
\end{array} has $ac=
3(-6)=-18
$ and $b=
7
.$
The two numbers with a product of $c$ and a sum of $b$ are $\left\{
9,-2
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
-4(3x^2+9x-2x-6)
.\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}
-4[(3x^2+9x)-(2x+6)]
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
-4[3x(x+3)-2(x+3)]
.\end{array}
Factoring the $GCF=
(x+3)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
-4[(x+3)(3x-2)]
\\\\=
-4(x+3)(3x-2)
.\end{array}
