If two zeroes of the polynomial p(x)
Question:

If two zeroes of the polynomial $p(x)=2 x^{4}-3 x^{3}-3 x^{2}+6 x-2$ are $\sqrt{2}$ and $-\sqrt{2}$, find its other two zeroes.

Solution:

Given: $\mathrm{p}(\mathrm{x})=2 x^{4}-3 x^{3}-3 x^{2}+6 x-2$ and the two zeroes, $\sqrt{2}$ and $-\sqrt{2}$

So, the polynomial is $(x+\sqrt{2})(x-\sqrt{2})=x^{2}-2$.

Let us divide $p(x)$ by $\left(x^{2}-2\right)$.

Here, $2 \mathrm{x}^{4}-3 \mathrm{x}^{3}-3 \mathrm{x}^{2}+6 \mathrm{x}-2=\left(x^{2}-2\right)\left(2 x^{2}-3 x+1\right)$

$=\left(x^{2}-2\right)\left[2 x^{2}-(2+1) x+1\right]$

$=\left(x^{2}-2\right)\left(2 x^{2}-2 x-x+1\right)$

$=\left(x^{2}-2\right)[(2 x(x-1)-1(x-1)]$

$=\left(x^{2}-2\right)(2 x-1)(x-1)$

$\therefore$ The other two zeroes are $\frac{1}{2}$ and 1