Answer
Remainder = 0,
Solution set =$\displaystyle \{-\frac{1}{2},1,2\}$
Work Step by Step
Divide $2x^{3}-5x^{2}+x+2\quad $with $x-2,\qquad (c=2)$
$\begin{array}{rrrrrrrr}
\underline{2}| &2 & -5 & 1 & 2 & & & & \\
& & 4 & -2 &-2 & & & & \\
& --& -- &-- &-- & & & & \\
& 2 & -1 & -1 & \fbox{0} & & & & \end{array}$
$\displaystyle \frac{dividend}{divisor}=quotient+\frac{remainder}{divisor}$, and the remainder is 0, so
$\displaystyle \frac{2x^{3}-5x^{2}+x+2}{(x-2)}=2x^{2}-x-1 $
... the trinomial on the RHS can be factored
(two factors of $-2$ with sum $-1$ are $+1$ and $-2$.
$2x^{2}-x-1 =2x^{2}-2x+x-1 =2x(x-1)+(x-1)$
$=(x-1)(2x+1)$
Thus,
$\displaystyle \frac{2x^{3}-5x^{2}+x+2}{(x-2)}=(x-1)(2x+1)$
$2x^{3}-5x^{2}+x+2=(x-2)(x-1)(2x+1)$
The equation $x^{3}-2x^{2}-x+2=0$ becomes
$(2x+1) (x-2)(x-1)=0\quad$ ... apply the zero product principle
Solution set =$\displaystyle \{-\frac{1}{2},1,2\}$
