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

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Answer 1 · 2016-02-07T23:21:44

$+-2i and -5$

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

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

so:
$(x^2+4)=0 or (x+5)^2=0$

The first part has not real solutions, since $(x^2+4)>0, AA x in RR$ ,

but has complex roots:

$(x^2+4)=0$
$x^2=-4$
$x=+-2i$

The second part is easy to resolve:

$(x+5)^2=0$

$x+5=0=> x=-5$

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