How do you factor #(p-5)^3+125#?

1 Answer
mason m
Jan 23, 2016

#p(p^2-15p+75)#

Explanation:

The sum of cubes identity can be written:

#a^3+b^3=(a+b)(a^2-ab+b^2)#

Here, we have

#(p-5)^3+125#

#=(p-5)^3+5^3#

#=((p-5)+5)((p-5)^2-(p-5)(5)+5^2)#

#=p(p^2-10p+25-5p+25+25)#

#=p(p^2-15p+75)#

The internal quadratic #p^2-15p+75# cannot be factored without using complex numbers.

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