Answer
The solutions of the equation are $ x=\left( -1,2,i,-i \right)$.
Work Step by Step
Use the rational root theorem from (1), ${{a}_{0}}=2$ and ${{a}_{n}}=1$.
The dividers of ${{a}_{0}}=1,2$ and ${{a}_{n}}=1$.
Now, as $\frac{{{a}_{0}}}{{{a}_{n}}}=\pm \frac{1,2}{1}$, we will check all the rational numbers.
At first we will check if $-1$ is a root of the expression.
Now, we will perform the synthetic division as below:
$\begin{matrix}
-1 & 1 & -1 & -1 & -1 & -2 \\
{} & {} & -1 & 2 & -1 & 2 \\
{} & 1 & -2 & 1 & -2 & 0 \\
\end{matrix}$
Since the remainder is zero, $ x=-1$ is a root of the given equation and so the equation ${{x}^{4}}-{{x}^{3}}-{{x}^{2}}-x-2=0$ can be factored as below:
$\begin{align}
& {{x}^{4}}-{{x}^{3}}-{{x}^{2}}-x-2=0 \\
& \left( x+1 \right)\left( {{x}^{3}}-2{{x}^{2}}+x-2 \right)=0 \\
& \left( x+1 \right)\left[ {{x}^{2}}\left( x-2 \right)+1\left( x-2 \right) \right]=0 \\
& \left( x+1 \right)\left( x-2 \right)\left( {{x}^{2}}+1 \right)=0
\end{align}$
Therefore, $ x=-1,2,+i,-i $
