# How do you use the Rational Root Theorem to find all the roots of 2x^3 – x^2 + 10x – 5 = 0 ?

Feb 4, 2016

The only Real root is $x = \frac{1}{2}$
Additional Complex roots $x = \pm i \sqrt{5}$

#### Explanation:

For the equation $\textcolor{b l u e}{2} {x}^{3} - {x}^{2} + 10 x - \textcolor{red}{5} = 0$
any rational roots that exist must be in the set:
color(white)("XXX"){+- ((color(red)("factors of " 5))/(color(blue)("factors of " 2)))}

$\textcolor{w h i t e}{\text{XXX}} = \left\{\pm \left(\frac{\textcolor{red}{1 , 5}}{\textcolor{b l u e}{1 , 2}}\right)\right\}$

$\textcolor{w h i t e}{\text{XXX}} = \left\{1 , \frac{1}{2} , 5 , \frac{5}{2} , - 1 , - \frac{1}{2} , - 5 , - \frac{5}{2}\right\}$

Using synthetic substitution with these candidate values: We find the only rational root is $x = \frac{1}{2}$
and that $2 {x}^{3} - {x}^{2} + 10 x - 5 = 0$ can be factored as
$\textcolor{w h i t e}{\text{XXX}} \left(x - \frac{1}{2}\right) \left(2 {x}^{2} + 10\right) = 0$

The $\left(x - \frac{1}{2}\right) = 0$ gave us the root $x = \frac{1}{2}$
but if $\left(2 {x}^{2} + 10\right) = 0$ is to give us any further roots:
$\textcolor{w h i t e}{\text{XXX}} 2 {x}^{2} = - 10$
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} = - 5$
$\textcolor{w h i t e}{\text{XXX}} x = \pm \sqrt{- 5}$
$\textcolor{w h i t e}{\text{XXX}} x = \pm i \sqrt{5}$