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How do you factor #(x^2-9)×(x^3-8)#?

1 Answer

Apr 9, 2015

#(x^2-9) = (x-2)(x+2)#

#(x^3-8) = (x-2)(x^2+2x+4)#
(note that #(x^2+2x+4)# has no Real factors; this may have been an intentional trick)

So
#(x^2-9)xx(x^3-8)#
can be factored as
#(x-2)^2(x+2)(x^2+2x+4)#

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