If 2 and –2 are two zeros of the polynomial

Solution:

Let $f(x)=2 x^{4}-5 x^{3}-11 x^{2}+20 x+12$

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

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

Therefore, one factor of $f(x)$ is $\left(x^{2}-4\right)$.

We get another factor of $f(x)$ by dividing it with $\left(x^{2}-4\right)$.

On division, we get the quotient $2 x^{2}-5 x-3$

$\Rightarrow f(x)=\left(x^{2}-4\right)\left(2 x^{2}-5 x-3\right)$

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

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

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

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

$\Rightarrow\left(x^{2}-4\right)(2 x+1)(x-3)=0$

$\Rightarrow\left(x^{2}-4\right)=0$ or $(2 x+1)=0$ or $(x-3)=0$

$\Rightarrow x=\pm 2,-\frac{1}{2}, 3$

Hence, all the zeroes of the polynomial $f(x)$ are $2,-2,-\frac{1}{2}$ and 3 .