The given polynomial is $f(x)=x^{4}+4 x^{3}-2 x^{2}-20 x-15$.

Since $(x-\sqrt{5})$ and $(x+\sqrt{5})$ are the zeroes of $f(x)$ it follows that each one of $(x-\sqrt{5})$ and $(x$ $+\sqrt{5})$ is a factor of $f(x)$

$(x-\sqrt{5})(x+\sqrt{5})=\left(x^{2}-5\right)$ is a factor of $f(x)$.

On dividing $f(x)$ by $\left(x^{2}-5\right)$, we get:

$\mathrm{f}(\mathrm{x})=0$

$$
\begin{aligned}
&\Rightarrow \mathrm{x}^{4}+4 \mathrm{x}^{3}-7 \mathrm{x}^{2}-20 \mathrm{x}-15=0 \\
&\Rightarrow\left(\mathrm{x}^{2}-5\right)\left(\mathrm{x}^{2}+4 \mathrm{x}+3\right)=0 \\
&\Rightarrow(\mathrm{x}-\sqrt{5})(\mathrm{x}+\sqrt{5})(\mathrm{x}+1)(\mathrm{x}+3)=0 \\
&\Rightarrow \mathrm{x}=\sqrt{5} \text { or } \mathrm{x}=-\sqrt{5} \text { or } \mathrm{x}=-1 \text { or } \mathrm{x}=-3
\end{aligned}
$$

Hence, all the zeroes are $\sqrt{5},-\sqrt{5},-1$ and $-3$.