Since $\sqrt{3}$ and $-\sqrt{3}$ are the zeroes of $f(x)$, it follows that each one of $(x-\sqrt{3})$ and $(x+\sqrt{3})$ is a factor of $\mathrm{f}(\mathrm{x})$

Consequently, $(x-\sqrt{3})(x+\sqrt{3})=\left(x^{2}-3\right)$ is a factor of $f(x)$.

On dividing $f(x)$ by $\left(x^{2}-3\right)$, we get:

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

$\Rightarrow\left(x^{2}+x-20\right)\left(x^{2}-3\right)=0$

$\Rightarrow\left(x^{2}+5 x-4 x-20\right)\left(x^{2}-3\right)$

$\Rightarrow[x(x+5)-4(x+5)]\left(x^{2}-3\right)$

$\Rightarrow(x-4)(x+5)(x-\sqrt{3})(x+\sqrt{3})=0$

$\Rightarrow \mathrm{x}=4$ or $\mathrm{x}=-5$ or $\mathrm{x}=\sqrt{3}$ or $\mathrm{x}=-\sqrt{3}$

Hence, all the zeroes are $\sqrt{3},-\sqrt{3}, 4$ and $-5$.