Obtain all other zeroes of $3x^{4} + 6x^{3} - 2x^{2} - 10x - 5$,if two of its zeroes are $\sqrt{\frac{5}{3}}$ and $-\sqrt{\frac{5}{3}}$.

(N/A) Let $p(x) = 3x^{4} + 6x^{3} - 2x^{2} - 10x - 5$.
Since $\sqrt{\frac{5}{3}}$ and $-\sqrt{\frac{5}{3}}$ are two zeroes of $p(x)$,then $(x - \sqrt{\frac{5}{3}})(x + \sqrt{\frac{5}{3}}) = (x^{2} - \frac{5}{3})$ is a factor of $p(x)$.
To find the other zeroes,we divide $p(x)$ by $(x^{2} - \frac{5}{3})$:
$3x^{4} + 6x^{3} - 2x^{2} - 10x - 5 = (x^{2} - \frac{5}{3})(3x^{2} + 6x + 3)$
$= 3(x^{2} - \frac{5}{3})(x^{2} + 2x + 1)$
$= 3(x^{2} - \frac{5}{3})(x + 1)^{2}$
Now,to find the remaining zeroes,we set $(x + 1)^{2} = 0$,which gives $x + 1 = 0$,so $x = -1$.
Since the factor is $(x + 1)^{2}$,the zero $x = -1$ is repeated twice.
Thus,the other two zeroes of the polynomial are $-1$ and $-1$.