# How do you find all the real and complex roots of (x^ 2 + 4)(x + 5)^2 = 0?

Feb 7, 2016

$\pm 2 i \mathmr{and} - 5$

#### Explanation:

(x^2+4)(x+5)^2=0

if a*b=0, then a=0 or b=0:

so:
$\left({x}^{2} + 4\right) = 0 \mathmr{and} {\left(x + 5\right)}^{2} = 0$

The first part has not real solutions, since $\left({x}^{2} + 4\right) > 0 , \forall x \in \mathbb{R}$,

but has complex roots:

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

The second part is easy to resolve:

${\left(x + 5\right)}^{2} = 0$

$x + 5 = 0 \implies x = - 5$