#### Answer

$f(x)=(x-1)(x+5)(x+4)$
Real Zeros: $-5,-4,1$ and all with multiplicity $1$.

#### Work Step by Step

Let us consider that $m$ is a factor of the constant term and $n$ is a factor of the leading coefficient. Then the potential zeros can be expressed by the possible combinations as: $\dfrac{m}{n}$.
We see from the given polynomial function that it has at most $3$ real zeros as degree is $3$.
The possible factors $m$ of the constant term and $n$ of the leading coefficient are: $m=\pm 1, \pm 2, \pm 3, \pm 6$ and $n=\pm 1$
Therefore, the possible rational roots of $f(x)$ are:
$\dfrac{m}{n}=\pm 1, \pm 2, \pm 3, \pm 6$
We test with synthetic division: we will try $x-1$.
$\left.\begin{array}{l}
1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
1 & 8 &11&-20\\\hline
&1&9 &20\\\hline
1&9 &20& |\ \ 0\end{array}$
Thus, we have: $f(x)=(x-1)(x^2+9x+20) \\=(x-1)(x+5)(x+4)$
Real Zeros: $-5,-4,1$ and all with multiplicity $1$.