# How do you find all the real and complex roots of x(x^2 - 4)(x^2 + 16) = 0?

Mar 5, 2016

$x = 0 , \pm 2 , \pm 4 i$

#### Explanation:

$x \left({x}^{2} - 4\right) \left({x}^{2} + 16\right) = 0$
$x \left(x + 2\right) \left(x - 2\right) \left(x + 4 i\right) \left(x - 4 i\right) = 0$
$\therefore$$x = 0 , \pm 2 , \pm 4 i$