# How do you find all the zeros of f(x)= 2x^3 + 3x^2+ 8x- 5?

Aug 3, 2016

$f \left(x\right)$ has zeros $\frac{1}{2}$ and $- 1 \pm 2 i$

#### Explanation:

$f \left(x\right) = 2 {x}^{3} + 3 {x}^{2} + 8 x - 5$

By the rational root theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 5$ and $q$ a divisor of the coefficient $2$ of the leading term.

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

$\pm \frac{1}{2} , \pm 1 , \pm \frac{5}{2} , \pm 5$

We find:

$f \left(\frac{1}{2}\right) = \frac{2}{8} + \frac{3}{4} + \frac{8}{2} - 5 = \frac{1}{4} + \frac{3}{4} + 4 - 5 = 0$

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

$2 {x}^{3} + 3 {x}^{2} + 8 x - 5$

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

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

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

$= \left(2 x - 1\right) \left(x + 1 - 2 i\right) \left(x + 1 + 2 i\right)$

Hence the other two zeros are:

$x = - 1 \pm 2 i$