Answer
x=2 is a solution because $f(2)=0$
Solution set = $\{-3,-1,2\}$
Work Step by Step
The graph suggests that $f(2)=0$. We divide f(x) with $(x-2)$, using synthetic division.
$\begin{array}{rrrrrrrr}
\underline{2}| & 1 & 2 & -5 &-6 & & & & \\
& & 2 & 8 &6 & & & & \\
& --& -- &-- &-- & & & & \\
& 1 & 4 & 3 & \fbox{0} & & & & \end{array}$
By the remainder theorem, $f(2)=0.$
$\displaystyle \frac{dividend}{divisor}=quotient+\frac{remainder}{divisor}\quad $with remainder =0,
$x^{3}+2x^{2}-5x-6=(x-2)(x^{2}+4x+3)$
... the trinomial on the RHS can be factored
(two factors of $3$ with sum $4$ are $+1$ and $+3$.
$x^{3}+2x^{2}-5x-6=(x-2)(x+1)(x+3)$
Solve
$(x-2)(x+1)(x+3)=0\quad $.... apply the zero product principle,
Solution set = $\{-3,-1,2\}$
