Find all the zeros of

Solution:

Let $f(x)=2 x^{4}-13 x^{3}+19 x^{2}+7 x-3$

It is given that $(2+\sqrt{3})$ and $(2-\sqrt{3})$ are two zeroes of $f(x)$

Thus, $f(x)$ is completely divisible by $(x-2-\sqrt{3})$ and $(x-2+\sqrt{3})$.

Therefore, one factor of $f(x)$ is $\left((x-2)^{2}-3\right)$

$\Rightarrow$ one factor of $f(x)$ is $\left(x^{2}-4 x+1\right)$

We get another factor of $f(x)$ by dividing it with $\left(x^{2}-4 x+1\right)$.

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

$\Rightarrow f(x)=\left(x^{2}-4 x+1\right)\left(2 x^{2}-5 x-3\right)$

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

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

$=(x-2-\sqrt{3})(x-2+\sqrt{3})(2 x+1)(x-3)$

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

$\Rightarrow(x-2-\sqrt{3})(x-2+\sqrt{3})(2 x+1)(x-3)=0$

$\Rightarrow(x-2-\sqrt{3})=0$ or $(x-2+\sqrt{3})=0$ or $(2 x+1)=0$ or $(x-3)=0$

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

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