How do you FOIL (2x - 3)(x^2 + 5x - 3)(2x3)(x2+5x3)?

2 Answers
Jun 3, 2015

This is where FOIL is not quite enough - It's meant for multiplying two binomials together, but here the second expression is a trinomial.

Instead you can break the problem into separate terms using the distributive law, multiply out, then recombine the terms...

(2x-3)(x^2+5x-3)(2x3)(x2+5x3)

=2x(x^2+5x-3)-3(x^2+5x-3)=2x(x2+5x3)3(x2+5x3)

=(2x^3+10x^2-6x)-(3x^2+15x-9)=(2x3+10x26x)(3x2+15x9)

=2x^3+10x^2-6x-3x^2-15x+9=2x3+10x26x3x215x+9

=2x^3+(10-3)x^2-(6+15)x+9=2x3+(103)x2(6+15)x+9

=2x^3+7x^2-21x+9=2x3+7x221x+9

Jun 3, 2015

Alternatively, working with each of the powers of xx from x^3x3 down to x^0x0 (i.e. the constant term), match the terms in the first and second bracketed expressions which will multiply to give that power of xx and add them together, thus:

Given: (2x-3)(x^2+5x-3)(2x3)(x2+5x3)

x^3 : 2x xx x^2 = 2x^3x3:2x×x2=2x3

x^2 : (2x xx 5x) + (-3 xx x^2) = 10x^2 - 3x^2 = 7x^2x2:(2x×5x)+(3×x2)=10x23x2=7x2

x^1 : (2x xx -3) + (-3 xx 5x) = -6x -15x = -21xx1:(2x×3)+(3×5x)=6x15x=21x

x^0 : -3 xx -3 = 9x0:3×3=9

Added, give 2x^3+7x^2-21x+92x3+7x221x+9