How do you FOIL #(x-2)(x+3)(x+1)# ?

1 Answer
George C.
Aug 13, 2017

#(x-2)(x+3)(x+1) = x^3+2x^2-5x-6#

Explanation:

FOIL only helps when multiplying one binomial by another, so will only allow us to do one of the multiplications. We can then use distributivity to complete the calculation...

#(x-2)(x+3)(x+1) = (x-2)(overbrace((x)(x))^"First"+overbrace((x)(1))^"Outside"+overbrace((3)(x))^"Inside"+overbrace((3)(1))^"Last")#

#color(white)((x-2)(x+3)(x+1)) = (x-2)(x^2+x+3x+3)#

#color(white)((x-2)(x+3)(x+1)) = (x-2)(x^2+4x+3)#

#color(white)((x-2)(x+3)(x+1)) = x(x^2+4x+3)-2(x^2+4x+3)#

#color(white)((x-2)(x+3)(x+1)) = (x^3+4x^2+3x)-(2x^2+8x+6)#

#color(white)((x-2)(x+3)(x+1)) = x^3+4x^2-2x^2+3x-8x-6#

#color(white)((x-2)(x+3)(x+1)) = x^3+2x^2-5x-6#

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