How do you find the complex roots of t^3+2t^2-4t-8=0?

Nov 26, 2016

The roots are:

$- 2 , - 2 , 2$

Explanation:

$0 = {t}^{3} + 2 {t}^{2} - 4 t - 8$

$\textcolor{w h i t e}{0} = \left({t}^{3} + 2 {t}^{2}\right) - \left(4 t + 8\right)$

$\textcolor{w h i t e}{0} = {t}^{2} \left(t + 2\right) - 4 \left(t + 2\right)$

$\textcolor{w h i t e}{0} = \left({t}^{2} - 4\right) \left(t + 2\right)$

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

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

So the roots are:

$- 2 , - 2 , 2$

There are no non-Real Complex roots.