How do you FOIL #(x+3)(x+3)(x+3)#?

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Answer 1 · 2015-06-16T23:03:27

FOIL will only get you part of the way, since once you multiply one pair of binomials you will have the product of a binomial and a trinomial, but

#(x+3)(x+3)(x+3) = x^3+9x^2+27x+27#

Using FOIL, first multiply #(x+3)(x+3)#

First: #x * x = x^2#
Outside: #x * 3 = 3x#
Inside: #3 * x = 3x#
Last: #3 * 3 = 9#

Add them together:
#F + O + I + L = x^2+3x+3x+9 = x^2+6x+9#

To multiply #(x^2+6x+9)(x+3)# use distributivity:

#(x^2+6x+9)(x+3)#

#=(x^2+6x+9)*x + (x^2+6x+9)*3#

#=(x^2*x)+(6x*x)+(9*x)+(x^2*3)+(6x*3)+(9*3)#

#=x^3+6x^2+9x+3x^2+18x+27#

#=x^3+(6x^2+3x^2)+(9x+18x)+27#

#=x^3+(6+3)x^2 + (9+18)x + 27#

#=x^3+9x^2+27x+27#

Alternatively, pick the 4th row of Pascal's triangle to get:

#1, 3, 3, 1#

List the first four ascending powers of three:

#1, 3, 9, 27#

Multiply these two sequences together to get:

#1, 9, 27, 27#

These are the coefficients of the descending powers of #x#:

#x^3+9x^2+27x+27#