#### Answer

$f(x)=2(x-2)(x+\sqrt 5)(x-\sqrt 5)$
Real Zeros: $-\sqrt 5, \sqrt 5,2$ 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 $1$ real zeros as degree is $3$.
The possible factors $m$ of the constant term and $n$ of the leading coefficient are: $m=\pm 1, \pm 2, \pm 5, \pm 10$and $n=\pm 1$
Therefore, the possible rational roots of $f(x)$ are:
$\dfrac{m}{n}=\pm 1, \pm 2, \pm 5, \pm 10$
We test with synthetic division; we will try $x-2$.
$\left.\begin{array}{l}
2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
1 & -2 &-5&10\\\hline
&2&0 &-10\\\hline
1&0 &-5& |\ \ 0\end{array}$
Thus, we have: $f(x)=2(x-2)(x^2-5)\implies f(x)=2(x-2)(x+\sqrt 5)(x-\sqrt 5)$
Real Zeros: $-\sqrt 5, \sqrt 5,2$ and all with multiplicity $1$.