#### Answer

$x=\left\{ -2,2,3 \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Express the given equation, $
x^3-3x^2-4x+12=0
,$ in factored form. Then use the Zero Product Property by equating each factor to zero. Finally, solve each of the resulting equations.
$\bf{\text{Solution Details:}}$
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
(x^3-3x^2)-(4x-12)=0
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
x^2(x-3)-4(x-3)=0
.\end{array}
Factoring the $GCF=
(x-3)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(x-3)(x^2-4)=0
.\end{array}
The expressions $
x^2
$ and $
4
$ are both perfect squares and are separated by a minus sign. Hence, $
x^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}
(x-3)(x+2)(x-2)=0
.\end{array}
Equating each factor to zero (Zero Product Property), then
\begin{array}{l}\require{cancel}
x-3=0
\\\\\text{OR}\\\\
x+2=0
\\\\\text{OR}\\\\
x-2=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
x-3=0
\\\\
x=3
\\\\\text{OR}\\\\
x+2=0
\\\\
x=-2
\\\\\text{OR}\\\\
x-2=0
\\\\
x=2
.\end{array}
Hence, the solutions are $
x=\left\{ -2,2,3 \right\}
.$