Answer
Remainder =0,
Solution set =$\displaystyle \{-2,-\frac{3}{2},\frac{1}{2}\}$
Work Step by Step
Divide $2x^{3}-3x^{2}-11x+6\quad $with $x+2,\qquad (c=-2)$
$\begin{array}{rrrrrrrr}
\underline{-2}| &2 & -3 & -11 &6 & & & & \\
& & -4 & 14 &-6 & & & & \\
& --& -- &-- &-- & & & & \\
& 2 & -7 & 3 & \fbox{0} & & & & \end{array}$
$\displaystyle \frac{dividend}{divisor}=quotient+\frac{remainder}{divisor}$, and the remainder is 0, so
$\displaystyle \frac{2x^{3}-3x^{2}-11x+6}{(x+2)}=2x^{2}-7x+3 $
... the trinomial on the RHS can be factored
(two factors of $ac=6$ with sum $b=-7$ are $-1$ and $-6$.
$2x^{2}-7x+3 =2x^{2}-x-6x+3 =x(2x-1)+3(2x-1)$
$=(2x-1)(2x+3)$
Thus,
$\displaystyle \frac{2x^{3}-3x^{2}-11x+6}{(x+2)}=(2x-1)(2x+3)$
$2x^{3}-3x^{2}-11x+6=(x+2)(2x-1)(2x+3)$
The equation $2x^{3}-3x^{2}-11x+6=0$ becomes
$(x+2)(2x-1)(2x+3)=0\quad$ ... apply the zero product principle
Solution set =$\displaystyle \{-2,-\frac{3}{2},\frac{1}{2}\}$
