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

Jun 16, 2015

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

$\left(x + 3\right) \left(x + 3\right) \left(x + 3\right) = {x}^{3} + 9 {x}^{2} + 27 x + 27$

Explanation:

Using FOIL, first multiply $\left(x + 3\right) \left(x + 3\right)$

First: $x \cdot x = {x}^{2}$
Outside: $x \cdot 3 = 3 x$
Inside: $3 \cdot x = 3 x$
Last: $3 \cdot 3 = 9$

$F + O + I + L = {x}^{2} + 3 x + 3 x + 9 = {x}^{2} + 6 x + 9$

To multiply $\left({x}^{2} + 6 x + 9\right) \left(x + 3\right)$ use distributivity:

$\left({x}^{2} + 6 x + 9\right) \left(x + 3\right)$

$= \left({x}^{2} + 6 x + 9\right) \cdot x + \left({x}^{2} + 6 x + 9\right) \cdot 3$

$= \left({x}^{2} \cdot x\right) + \left(6 x \cdot x\right) + \left(9 \cdot x\right) + \left({x}^{2} \cdot 3\right) + \left(6 x \cdot 3\right) + \left(9 \cdot 3\right)$

$= {x}^{3} + 6 {x}^{2} + 9 x + 3 {x}^{2} + 18 x + 27$

$= {x}^{3} + \left(6 {x}^{2} + 3 {x}^{2}\right) + \left(9 x + 18 x\right) + 27$

$= {x}^{3} + \left(6 + 3\right) {x}^{2} + \left(9 + 18\right) x + 27$

$= {x}^{3} + 9 {x}^{2} + 27 x + 27$

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 $x$:

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