#### Answer

$(w+2)(w-2)(w+3)$

#### Work Step by Step

Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
w^3-4w+3w^2-12
\\\\=
(w^3-4w)+(3w^2-12)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
w(w^2-4)+3(w^2-4)
.\end{array}
Factoring the $GCF=
(w^2-4)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(w^2-4)(w+3)
.\end{array}
The expressions $
w^2
$ and $
4
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
w^2-4
,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
[(w)^2-(2)^2](w+3)
\\\\=
[(w+2)(w-2)](w+3)
\\\\=
(w+2)(w-2)(w+3)
.\end{array}