## Precalculus (6th Edition) Blitzer

The solutions of the equation are $x=\left( -1,2,i,-i \right)$.
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, $x=-1,2,+i,-i$