# How do you find the rest of the zeros given one of the zero c=1/2 and the function f(x)=2x^3-x^2-10x+5?

Apr 17, 2017

$\sqrt{5}$ and $- \sqrt{5}$ are other two zeros of $f \left(x\right) = 2 {x}^{3} - {x}^{2} - 10 x + 5$.

#### Explanation:

As $\frac{1}{2}$ is a zero of function $f \left(x\right) = 2 {x}^{3} - {x}^{2} - 10 x + 5$, $\left(x - \frac{1}{2}\right)$ is a factor of $f \left(x\right)$. Note that coefficient of ${x}^{3}$ is $2$, hence one can say $2 \left(x - \frac{1}{2}\right) = 2 x - 1$ is a factor.

We can easily divide $f \left(x\right) = 2 {x}^{3} - {x}^{2} - 10 x + 5$ by $2 x - 1$ as follows:

$f \left(x\right) = 2 {x}^{3} - {x}^{2} - 10 x + 5$

= ${x}^{2} \left(2 x - 1\right) - 5 \left(2 x - 1\right)$

= $\left(2 x - 1\right) \left({x}^{2} - 5\right)$

It is easy to factorize ${x}^{2} - 5$ using ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$ and

${x}^{2} - 5 = {x}^{2} - {\left(\sqrt{5}\right)}^{2} = \left(x + \sqrt{5}\right) \left(x - \sqrt{5}\right)$

And hence $\left(x + \sqrt{5}\right)$ and $\left(x - \sqrt{5}\right)$ are also factors of $f \left(x\right)$ and therefore

$\sqrt{5}$ and $- \sqrt{5}$ are other two zeros of $f \left(x\right) = 2 {x}^{3} - {x}^{2} - 10 x + 5$.