Answer
$6(x-2)(x+1)$
Work Step by Step
Factoring the $GCF=
6
$ results to $
6(x^2-x-2)
$. The two numbers whose product is $
-2
$ and whose sum is $
-1
$ are $\{
-2,1
\}$. Using these numbers to decompose the middle term of the trinomial results to
\begin{array}{l}
6(x^2-x-2)
\\=
6(x^2-2x+x-2)
\\=
6[(x^2-2x)+(x-2)]
\\=
6[x(x-2)+(x-2)]
\\=
6[(x-2)(x+1)]
\\=
6(x-2)(x+1)
.\end{array}
