# How do you find the roots of the polynomial -x^3+5x^2-11x+55=0?

Jan 12, 2017

$x = 5 \text{ }$ or $\text{ } x = \pm \sqrt{11} i$

#### Explanation:

This cubic factors by grouping:

$0 = - {x}^{3} + 5 {x}^{2} - 11 x + 55$

$\textcolor{w h i t e}{0} = \left(- {x}^{3} + 5 {x}^{2}\right) + \left(- 11 x + 55\right)$

$\textcolor{w h i t e}{0} = - {x}^{2} \left(x - 5\right) - 11 \left(x - 5\right)$

$\textcolor{w h i t e}{0} = - \left({x}^{2} + 11\right) \left(x - 5\right)$

$\textcolor{w h i t e}{0} = - \left({x}^{2} - {\left(\sqrt{11} i\right)}^{2}\right) \left(x - 5\right)$

$\textcolor{w h i t e}{0} = - \left(x - \sqrt{11} i\right) \left(x + \sqrt{11} i\right) \left(x - 5\right)$

Hence:

$x = 5 \text{ }$ or $\text{ } x = \pm \sqrt{11} i$