How do you factor completely #-a^3-100a#?

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Answer 1 · 2016-06-28T09:37:26

#-a^3-100a=-a(a^2+100)=-a(a-10i)(a+10i)#

#-a^3-100a#

= #-a(a^2+100)#

If we domain is real numbers, we cannot factorize it further.

However, if domain is complex numbers above is equal to

#-a[a^2-10ai+10ai-100i^2]#

= #-a[a(a-10i)+10i(a-10i)]#

= #-a(a-10i)(a+10i)#