The given polynomial is $f(x)=2 x^{4}-3 x^{3}-5 x^{2}+9 x-3$

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:

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

$\Rightarrow 2 x^{4}-3 x^{3}-5 x^{2}+9 x-3=0$

$\Rightarrow\left(x^{2}-3\right)\left(2 x^{2}-3 x+1\right)=0$

$\Rightarrow\left(x^{2}-3\right)\left(2 x^{2}-2 x-x+1\right)=0$

$\Rightarrow(x-\sqrt{3})(x+\sqrt{3})(2 x-1)(x-1)=0$

$\Rightarrow \mathrm{x}=\sqrt{3}$ or $\mathrm{x}=-\sqrt{3}$ or $\mathrm{x}=\frac{1}{2}$ or $\mathrm{x}=1$

Hence, all the zeroes are $\sqrt{3},-\sqrt{3}, \frac{1}{2}$ and 1