Find all the zeroes of $2x^{4}-3x^{3}-3x^{2}+6x-2$,if you know that two of its zeroes are $\sqrt{2}$ and $-\sqrt{2}$.

(N/A) Since two zeroes are $\sqrt{2}$ and $-\sqrt{2}$,$(x-\sqrt{2})(x+\sqrt{2}) = x^{2}-2$ is a factor of the given polynomial.
Now,we divide the given polynomial by $x^{2}-2$ to find the other factors.
Performing the division:
$2x^{4}-3x^{3}-3x^{2}+6x-2 = (x^{2}-2)(2x^{2}-3x+1)$
Now,we factorise the quadratic polynomial $2x^{2}-3x+1$ by splitting the middle term:
$2x^{2}-2x-x+1 = 2x(x-1)-1(x-1) = (2x-1)(x-1)$
Setting these factors to zero,we get:
$2x-1 = 0 \implies x = \frac{1}{2}$
$x-1 = 0 \implies x = 1$
Therefore,all the zeroes of the given polynomial are $\sqrt{2}, -\sqrt{2}, \frac{1}{2},$ and $1$.