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.