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#