#### Answer

$\{ -3,\ -1,\ 2 \}$

#### Work Step by Step

The graph suggests that x=2 is a solution to the equation.
We check by synthetic division(the last number in the last row is $f(2)$
Also, if it is a zero, then, by the Factor Theorem, $(x-2)$ is a factor of f...
$(x^{3}+2x^{2}-5x-6)\div(x-2)$
$\begin{array}{lllllll}
\underline{2}| & 1 & 2 & -5 & -6 & & \\
& & 2 & 8 & 6 & & \\
& -- & -- & -- & -- & -- & \\
& 1 & 4 & 3 & |\overline{ 0 } & &
\end{array}$
So, $(x-2)$ is a factor of $f(x).$
$f(x)=(x-2)(x^{2}+4x+3)$
we try to factor the trinomial by searching for two integers
whose product is $3$ and whose sum is $4$....
We find $1$ and $3:\quad x^{2}+4x+3=(x+1)(x+3)$
So, f(x) has three factors of form $(x-c)$,
$f(x)=(x-2)(x+1)(x+3)$,
and by (b) of the Factor Theorem,
zeros of f: $ -3,\ -1,\ 2$
make up the solution set of the equation $f(x)=0:$
$\{ -3,\ -1,\ 2 \}$