# How do you find the complex roots of n^3-9n=0?

Sep 25, 2017

$\text{there are no complex roots}$

#### Explanation:

${n}^{3} - 9 n = 0$

$\Rightarrow n \left({n}^{2} - 9\right) = 0$

$\Rightarrow n \left(n - 3\right) \left(n + 3\right) = 0$

$\text{equating each factor to zero and solving for n}$

$\Rightarrow n = 0$

$n - 3 = 0 \Rightarrow n = 3$

$n + 3 = - \Rightarrow n = - 3$

$\text{the roots "n=0,n=3,n=-3" are real}$