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

Two zeroes are $$\sqrt{{5}\over{3}}$$ and $$-\sqrt{{5}\over{3}}$$.

$$=> (x -\sqrt{{5}\over{3}}) ( x + \sqrt{{5}\over{3}}) = x ^2 - {{5}\over {3}}$$
= $${{1}\over {3}} × (3x^2 - 5)$$

Both $${{1}\over {3}}$$ and $$(3x^2 - 5)$$ are the factors of $$3x^4 + 6x^3 - 2x^2 -10x - 5$$.

So, we'll divide the given polynomial by $$3x^2 - 5$$

Solving it using long division:

$$\begin{array}{rrrr|ll} 6x^3 + 3x^4 & -2x^2 & -10x & -5 & 3x^2 - 5 \\ -3x^4 & +5x^2 & & & x^2 + 2x + 1 \\ \hline 6x^3 & +3x^2 & -10x & -5\\ \phantom{-}-6x^3 & & + 10x & & & \\ \hline & 3x^2 & & -5 \\ & -3x^2 & & +5 \\ \hline & & & 0\\ \hline \end{array}$$
$$3x^4 + 6x^3 - 2x^2 -10x - 5$$
=> $$(3x^2 - 5) (x^2 + 2x + 1)$$
=> $$(3x^2 - 5) (x(x + 1) + 1(x + 1))$$
=> $$(3x^2 - 5) (x + 1)(x + 1)$$
So the other two zeroes are x = -1 and -1.