#### Answer

Remainder = 0,
Solution set =$\{-1,1,2\}$

#### Work Step by Step

Divide $x^{3}-2x^{2}-x+2$
with $x+1,\qquad (c=-1)$
$\begin{array}{rrrrrrrr}
\underline{-1}| &1 & -2 & -1 & 2 & & & & \\
& & -1 & 3 &-2 & & & & \\
& --& -- &-- &-- & & & & \\
& 1 & -3 & 2 & \fbox{0} & & & & \end{array}$
$\displaystyle \frac{dividend}{divisor}=quotient+\frac{remainder}{divisor}$, and the remainder is 0, so
$\displaystyle \frac{x^{3}-2x^{2}-x+2}{(x+1)}=x^{2}-3x+2 $
... the trinomial on the RHS can be factored
(two factors of $2$ with sum $-3$ are $-1$ and $-2$.
Thus, $ x^{2}-3x+2 =(x-2)(x-1)\quad$ and
$x^{3}-2x^{2}-x+2=(x+1)(x-1)(x-2)$
The equation $x^{3}-2x^{2}-x+2=0$ becomes
$(x+1)(x-1)(x-2) =0\quad$ ... apply the zero product principle
Solution set =$\{-1,1,2\}$