#### Answer

$x=\left\{ -2,0,2 \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
-x^3-3x^2+4x+12=0
,$ express it first in factored form. Then equate each factor to zero using the Zero Product Property. Finally, solve each resulting equation.
$\bf{\text{Solution Details:}}$
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
\\\\
(x+3)(-1)(x^2-4)=0
\\\\
\dfrac{(x+3)(-1)(x^2-4)}{-1}=\dfrac{0}{-1}
\\\\
(x+3)(x^2-4)=0
.\end{array}
The expressions $
x^2
$ and $
4
$ are both perfect squares (the square root is exact) 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), the solutions to the equation above are
\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, $
x=\left\{ -2,0,2 \right\}
.$