Answer
The roots are $x=-3$ , $x=4$ , $x=i$ ,and $x=-i$.
Work Step by Step
Consider the quadratic function ${{x}^{4}}-{{x}^{3}}-11{{x}^{2}}=x+12$.
By the trial and error method, it is observed that $x=4$ is a root of the function. Thus $\left( x-4 \right)$ is a factor. By long division:
$x-4\overset{{{x}^{3}}+3{{x}^{2}}+x+3}{\overline{\left){\begin{align}
& {{x}^{4}}-{{x}^{3}}-11{{x}^{2}}-x-12 \\
& \underline{{{x}^{4}}-4{{x}^{3}}} \\
& 3{{x}^{3}}-11{{x}^{2}}-x-12 \\
& \underline{3{{x}^{3}}-12{{x}^{2}}} \\
& {{x}^{2}}-x-12 \\
& \underline{{{x}^{2}}-4x} \\
& 3x-12 \\
& \underline{3x-12} \\
& 0 \\
\end{align}}\right.}}$
Thus, the function can be expressed as $\left( x-4 \right)\left( {{x}^{3}}+3{{x}^{2}}+x+3 \right)=0$
Simplify further:
$\begin{align}
& \left( x-4 \right)\left( {{x}^{3}}+3{{x}^{2}}+x+3 \right)=0 \\
& \left( x-4 \right)\left( {{x}^{2}}\left( x+3 \right)+1\left( x+3 \right) \right)=0 \\
& \left( x-4 \right)\left( x+3 \right)\left( {{x}^{2}}+1 \right)=0 \\
& x=4,-3,\pm i
\end{align}$
Thus, the roots of the polynomial of the equation ${{x}^{4}}-{{x}^{3}}-11{{x}^{2}}-x-12=0$ are $x=-3$ , $x=4$ , $x=i$ ,and $x=-i$.