# How do you find all the real and complex roots of x^3 + 8 = 0?

Jan 12, 2016

Roots: x in {-2, color(white)("X")1+sqrt(3)i, color(white)("X")1-sqrt(3)i)}

#### Explanation:

Since ${\left(- 2\right)}^{3} = - 8$
$\textcolor{w h i t e}{\text{XXX}} x = \left(- 2\right)$ is an obvious root of $\left({x}^{3} + 8 = 0\right)$

Therefore $\left(x + 2\right)$ is a factor of $\left({x}^{3} + 8\right)$

$\left({x}^{3} + 8\right) \div \left(x + 2\right) = {x}^{2} - 2 x + 4$
$\textcolor{w h i t e}{\text{XXX}}$use synthetic or polynomial long division to get
$\textcolor{w h i t e}{\text{XXX}}$this result if it is not apparent.

$\textcolor{w h i t e}{\text{XXX}} x = \frac{- b \pm \sqrt{{b}^{2} + 4 a c}}{2 a}$
with $a = 1 , b = \left(- 2\right) , c = 4$
$\textcolor{w h i t e}{\text{XXX}} x = 1 \pm \sqrt{3} i$