Question

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

Answer 1

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`

Explanation:

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`