Obtain all the zeros of the polynomial

Solution:

Let $f(x)=x^{4}+x^{3}-14 x^{2}-2 x+24$

It is given that $\sqrt{2}$ and $-\sqrt{2}$ are two zeroes of $f(x)$

Thus, $f(x)$ is completely divisible by $(x+\sqrt{2})$ and $(x-\sqrt{2})$.

Therefore, one factor of $f(x)$ is $\left(x^{2}-2\right)$.

We get another factor of $f(x)$ by dividing it with $\left(x^{2}-2\right)$.

On division, we get the quotient $x^{2}+x-12$

$\Rightarrow f(x)=\left(x^{2}-2\right)\left(x^{2}+x-12\right)$

$=\left(x^{2}-2\right)\left(x^{2}+4 x-3 x-12\right)$

$=\left(x^{2}-2\right)(x(x+4)-3(x+4))$

$=\left(x^{2}-2\right)(x+4)(x-3)$

To find the zeroes, we put $f(x)=0$

$\Rightarrow\left(x^{2}-2\right)(x+4)(x-3)=0$

$\Rightarrow\left(x^{2}-2\right)=0$ or $(x+4)=0$ or $(x-3)=0$

$\Rightarrow x=\pm \sqrt{2},-4,3$

Hence, all the zeroes of the polynomial $f(x)$ are $\sqrt{2},-\sqrt{2},-4$ and 3 .