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 $x$ :

$x^3+9x^2+27x+27$

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