Answer
$f(x)=(x+1)(x+3)(x-2)$
Zeros: $-3,\ \ -1,\ \ 2,$
all with multiplicity 1.
Work Step by Step
$f(x)$ has at most 3 real zeros, as the degree is 3.
Rational zeros: $\displaystyle \frac{p}{q},$ in lowest terms,
$p$ must be a factor of $-6,\qquad \pm 1,\pm 2,\pm 3,\pm 6$
and $q$ must be a factor of $ 1,\qquad \pm 1$
$\displaystyle \frac{p}{q}=\pm 1,\pm 2,\pm 3,\pm 6$
Testing with synthetic division, try $x+1$
$\left.\begin{array}{l}
-1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
1 &2 &-5&-6\\\hline
&-1&-1 &6\\\hline
1&1 &-6& |\ \ 0\end{array}$
$f(x)=(x+1)(x^{2}+x-6)$
... factoring, we need factors of -6 with sum=1... they are +3 and -2
$f(x)=(x+1)(x+3)(x-2)$
Zeros: $-3,\ \ -1,\ \ 2,$
all with multiplicity 1.