Answer
$f(x)=(x-1)(x+5)(x+4)$
Zeros: $-5,\ \ -4,\ \ 1,$
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 $-20,\qquad \pm 1,\pm 2,\pm 4,\pm 5,\pm 10,\pm 20$
and $q$ must be a factor of $ 1,\qquad \pm 1$
$\displaystyle \frac{p}{q}=\pm 1,\pm 2,\pm 4,\pm 5,\pm 10,\pm 20$
Testing with synthetic division, try $x-1$
$\left.\begin{array}{l}
1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
1 &8 &11 &-20\\\hline
&1 &9 &20 \\\hline
1&9 &20& |\ \ 0 \end{array}$
$f(x)=(x-1)(x^{2}+9x+20)$
... factoring, we need factors of $20$ with sum=$4$1... they are $+5$ and $+4.$
$f(x)=(x-1)(x+5)(x+4)$
Zeros: $-5,\ \ -4,\ \ 1,$
all with multiplicity 1.