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

Given,  zeroes are $\sqrt{\frac{5}{3}}$ and $-\sqrt{\frac{5}{3}}$.

Then, $\left(x-\sqrt{\frac{5}{3}}\right)\left(x+\sqrt{\frac{5}{3}}\right)$ is a factor.

$\left(x-\sqrt{\frac{5}{3}}\right)\left(x+\sqrt{\frac{5}{3}}\right)=x^2-\frac{5}{3}.$

For finding the other factor, we have to divide the given polynomial by $x^2-\frac{5}{3}.$

$$\begin{gathered} \begin{array}{ccccc}3x^2& +& 6x& +3& \end{array} \\ \begin{array}{ccc}{x}^2& -& \frac{5}{3}\end{array}\sqrt{\begin{array}{ccccc}3x^4& + 6x^3& - 2x^2& - 10x& - 5\\ 3x^4& & - 5x^2& & \\ (-) & & (+) & & \end{array}} \\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ \begin{array}{cccc}6x^3& + 3x^2& - 10x& - 5\\ 6x^3& & - 10x& \\ (-) & & (+) & \end{array} \\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ \begin{array}{cc}3x^2& - 5\\ 3x^2& - 5\\ (-) & (+) \end{array} \\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ 0 \end{gathered}$$

So, $3x^2+6x+3$ is the other factor.

$3x^2+6x+3=3x^2+3x+3x+3=3{(x+1)}^2.$

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

Therefore, the other two zeros are -1 and -1.