# How do you find all the zeros of f(x)=2x^3-x^2-3x-1?

Aug 2, 2016

Zeros: $- \frac{1}{2}$ and $\frac{1}{2} \pm \frac{\sqrt{5}}{2}$

#### Explanation:

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

By the rational root theorem any rational zeros of $f \left(x\right)$ must be expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 1$ and $q$ a divisor of the coefficient $2$ of the leading term.

That means that the only possible rational zeros of $f \left(x\right)$ are:

$\pm \frac{1}{2} , \pm 1$

We find:

$f \left(- \frac{1}{2}\right) = 2 \left(- \frac{1}{8}\right) - \left(\frac{1}{4}\right) - 3 \left(- \frac{1}{2}\right) - 1 = - \frac{1}{4} - \frac{1}{4} + \frac{3}{2} - 1 = 0$

So $x = - \frac{1}{2}$ is a zero and $\left(2 x + 1\right)$ a factor:

$2 {x}^{3} - {x}^{2} - 3 x - 1$

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

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

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

$= \left(2 x + 1\right) \left(x - \frac{1}{2} - \frac{\sqrt{5}}{2}\right) \left(x - \frac{1}{2} + \frac{\sqrt{5}}{2}\right)$

Hence the other two zeros are:

$x = \frac{1}{2} \pm \frac{\sqrt{5}}{2}$