# Find all the zeros of

Question:

Find all the zeros of $2 x^{4}-3 x^{3}-3 x^{2}+6 x-2$ if it is given that two of its zeros are 1 and $\frac{1}{2}$.

Solution:

Let $f(x)=2 x^{4}-3 x^{3}-3 x^{2}+6 x-2$

It is given that 1 and $\frac{1}{2}$ are two zeroes of $f(x)$.

Thus, $f(x)$ is completely divisible by $(x-1)$ and $\left(x-\frac{1}{2}\right)$.

Therefore, one factor of $f(x)$ is $(x-1)\left(x-\frac{1}{2}\right)$

$\Rightarrow$ one factor of $f(x)$ is $\left(x^{2}-\frac{3}{2} x+\frac{1}{2}\right)$

We get another factor of $f(x)$ by dividing it with $\left(x^{2}-\frac{3}{2} x+\frac{1}{2}\right)$.

On division, we get the quotient $2 x^{2}-4$

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

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

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

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

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

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

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