How do you factor #w^3-3w^2-9w+27#?

1 Answer
Sep 26, 2017

Answer:

#(w^3-3w^2-9w+27)=color(blue)((x-3)(x-3)(x+3))#

Explanation:

Writing #w^3-3w^2-9w+27#
as
#color(white)("XXX")w^3-3w^2-3^3w+3^3#
makes it fairly clear that
#color(white)("XXX")w=3# is a zero of this expression.

That is #(w-3)# is a factor of #(w^3-3w^2-9w+27)#

If we divided #(w^3-3w^2-9w+27)# by #(w-3)#
(either by polynomial long division or synthetic division)
we find
#color(white)("XXX")(x^2-9)# is also a factor

...but #(x^2-9)=(x^2-3^2)=(x-3)(x+3)#

So a complete factoring is
#color(white)("XXX")(w^3-3w^2-9w+27)=(x-3)(x-3)(x+3)#