Answer
By the remainder theorem, $f(-3)=0.$
Solution set = $\displaystyle \{-3,\frac{1}{2},2\}$
Work Step by Step
The graph suggests that $f(-3)=0$. We divide f(x) with $(x+3)$, using synthetic division.
$\begin{array}{rrrrrrrr}
\underline{-3}| & 2 & 1 & -13 & 6 & & & & \\
& & -6 & 15 & -6 & & & & \\
& --& -- &-- &-- & & & & \\
& 2 & -5 & 2 & \fbox{0} & & & & \end{array}$
By the remainder theorem, $f(-3)=0.$
$\displaystyle \frac{dividend}{divisor}=quotient+\frac{remainder}{divisor}\quad $with remainder =0,
$2x^{3}+x^{2}-13x+6=(x+3)(2x^{2}-5x+2)$
... the trinomial on the RHS can be factored
(two factors of $ac=4$ with sum $-5$ are $-1$ and $-4$.
$2x^{2}-5x+2=2x^{2}-4x-x+2=2x(x-2)-(x-2)$
$=(x-2)(2x-1)$
Thus,
$2x^{3}+x^{2}-13x+6=(x+3)(x-2)(2x-1)$
Solve
$(x+3)(x-2)(2x-1)=0\quad $.... apply the zero product principle,
Solution set = $\displaystyle \{-3,\frac{1}{2},2\}$
