Answer
$x\displaystyle \in\{-3,-i,i, 4\}$
Work Step by Step
See The Rational Zero Theorem:
If $\frac{p}{q}$ is a zero of the polynomial $f(x) $with integer coefficients,
then $p$ is a factor of the constant term, $a_{0}$, and
$q$ is a factor of the leading coefficient, $a_{n}$.
------------------------
$x^4-x^{3}-11x^{2}-x-12=0$
a. Candidates for zeros, $\displaystyle \frac{p}{q}:$
$p:\qquad \pm 1, \pm 2, \pm 3, \pm4, \pm6, \pm12$
$q:\qquad \pm 1$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2,\pm 3, \pm4, \pm6, \pm12$
b. Try for $x=4:$
$\begin{array}{lllll}
\underline{4}| & 1 & -1 & -11 & -1& -12\\
& & 4 & 12 & 4& 12\\
& -- & -- & -- & --\\
& 1 & 3 & 1& 3 & |\underline{0}
\end{array}$
$4$ is a zero,
$(x-4)(x^{3} +3x^2+x+3)=0$
c. Factorize the Quadinomial $(x^3+3x^2+x+3)$ using distributive property of polynomial function.
$x^{3}+3x^2+x+3=x^2(x+3)+1(x+3)$
$(x^2+1)(x+3),$
$(x-4)(x^2+1)(x+3)=0$
$x\displaystyle \in\{-3,-i,i, 4\}$
