How do you use FOIL to multiply #(x-2) (x^2-5x+3)#?

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Answer 1 · 2015-06-17T11:19:32

FOIL helps with multiplying two binomials, but is not applicable to multiplying a binomial by a trinomial.

Instead use distributivity to find:

#(x-2)(x^2-5x+3) = x^3-7x^2+13x-6#

#(x-2)(x^2-5x+3)#

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

#=x^3-5x^2+3x-2x^2+10x-6#

#=x^3-(5x^2+2x^2)+(3x+10x)-6#

#=x^2-(5+2)x^2+(3+10)x-6#

#=x^2-7x^2+13x-6#

Alternatively, do what I do, look at each of the powers of #x# in descending order in turn and total up the coefficients:

#x^3# : #1*1 = 1#
#x^2# : #(1*-5)+(-2*1) = -5-2 = -7#
#x# : #(1*3)+(-2*-5) = 3+10 = 13#
#1# : #-2*3 = -6#

Putting these together:

#x^3-7x^2+13x-6#