How do you use FOIL to multiply #(x-2)^2 (x+17)#?

1 Answer
George C.
Jun 23, 2015

FOIL will only get you part of the way, but...

#(x-2)^2(x+17)#

#= (x^2-4x+4)(x+17)#

#= x^3+13x^2-64x+68#

Explanation:

FOIL will only get you part of the way since once you multiply two of the binomials together you will be left with multiplying a binomial by a trinomial - which is not covered by FOIL.

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

#= F + O + I + L# (First, Outside, Inside, Last)

#= (x*x)+(x*-2)+(-2*x)+(-2*-2)#

#=x^2-2x-2x+4#

#= x^2-4x+4#

Then using distributivity...

#(x^2-4x+4)(x+17)#

#=(x^2-4x+4)x + (x^2-4x+4)17#

#=x^3-4x^2+4x+17x^2-68x+68#

#=x^2+(17-4)x^2-(68-4)x+68#

#=x^2+13x^2-64x+68#

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