**Question:**

Find other zeroes of the polynomial $p(x)=2 x^{4}+7 x^{3}-19 x^{2}-14 x+30$ if two of its zeroes are $\sqrt{2}$ and $\sqrt{-2}$.

**Solution:**

The given polynomial is $p(x)=2 x^{4}+7 x^{3}-19 x^{2}-14 x+30$ and two zeroes of this polynomial are $-\sqrt{2}$ and $\sqrt{2}$, then we have to find the two other zeroes of $p(x)$ If two zeroes of the polynomial $p(x)$ are $-\sqrt{2}$ and $\sqrt{2}$, then

$(x+\sqrt{2})(x-\sqrt{2})=x^{2}-2$ is a factor of $p(x)$

Therefore can be written as

$p(x)=2 x^{2}\left(x^{2}-2\right)+7 x\left(x^{2}-2\right)-15\left(x^{2}-2\right)$

$=\left(x^{2}-2\right)\left(2 x^{2}+7 x-15\right)$

The other two zeroes are obtained from the polynomial

$2 x^{2}+7 x-15=2 x^{2}-10 x-3 x-15$

$=(2 x-3)(x+5)$

$\Rightarrow x=\frac{3}{2}$ and $x=-5$

Hence the other zeroes are $\frac{3}{2}$ and $-5$