Basic FOIL

Recall the the distributive law: for all real numbers $\,a\,$, $\,b\,$, and $\,c\,$, $\,a(b+c) = ab + ac\,$.

At first glance, it might not look like the distributive law applies to the expression $\,(a+b)(c+d)\,$.
However, it does—once you apply a popular mathematical technique called ‘treat it as a singleton’.
Here's how it goes:

First, rewrite the distributive law using some different variable names: $\,z(c+d) = zc + zd\,$.
This says that anything times $\,(c+d)\,$ is the anything times $\,c\,$, plus the anything times $\,d\,$.

Now, look back at $\,(a+b)(c+d)\,$, and take the group $\,(a+b)\,$ as $\,z\,$.
That is, you're taking something that seems to have two parts,
and you're treating it as a single thing, a ‘singleton’!
Look what happens:

$(a+b)(c+d)$
	$= \overset{z}{\overbrace{(a+b)}}(c+d)$	(give $\,(a+b)\,$ the name $\,z\,$)
	$= z(c + d)$	(rewrite)
	$= zc + zd$	(use the distributive law)
	$= (a+b)c + (a+b)d$	(since $\,z = a + b\,$)
	$= ac + bc + ad + bd$	(use the distributive law twice)
	$= ac + ad + bc + bd$	(re-order; switch the two middle terms)
	$= \underset{\text{F}}{\underbrace{\ ac\ }} + \underset{\text{O}}{\underbrace{\ ad\ }} + \underset{\text{I}}{\underbrace{\ bc\ }} + \underset{\text{L}}{\underbrace{\ bd\ }}$

You get four terms, and each of these terms is assigned a letter.
These letters form the word FOIL ,
and provide a powerful memory device
for multiplying out expressions of the form $\,(a+b)(c+d)\,$.

Here's the meaning of each letter in the word FOIL:

The first number in the group $\,(a+b)\,$ is $\,a\,$;
the first number in the group $\,(c+d)\,$ is $\,c\,$.
Multiplying these Firsts together gives $\,ac\,$, which is labeled ‘ F ’.
When you look at the expression $\,(a+b)(c+d)\,$ from far away,
you see $\,a\,$ and $\,d\,$ on the outside.
That is, $\,a\,$ and $\,d\,$ are the outer numbers.
Multiplying these Outers together gives $\,ad\,$, which is labeled ‘ O ’.
Similarly, when you look at the expression $\,(a+b)(c+d)\,$ from far away,
you see $\,b\,$ and $\,c\,$ on the inside.
That is, $\,b\,$ and $\,c\,$ are the inner numbers.
Multiplying these Inners together gives $\,bc\,$, which is labeled ‘ I ’.
The last number in the group $\,(a+b)\,$ is $\,b\,$;
the last number in the group $\,(c+d)\,$ is $\,d\,$.
Multiplying these Lasts together gives $\,bd\,$, which is labeled ‘ L ’.

One common application of FOIL is to multiply out expressions like $\,(x-1)(x+4)\,$.
Remember the exponent laws, and be sure to combine like terms whenever possible:

$\,(x-1)(x+4)$

$= \underset{\text{F}}{\underbrace{(x\cdot x)}} + \underset{\text{O}}{\underbrace{(x\cdot 4)}} + \underset{\text{I}}{\underbrace{(-1\cdot x)}} + \underset{\text{L}}{\underbrace{(-1\cdot 4)}} $

$= x^2 + 4x - x - 4$

$= x^2 + 3x - 4\,$

You want to be able to write this down without including the first step above:

$(x-1)(x+4) = \underset{\text{F}}{\underbrace{\ x^2\ }} + \underset{\text{O}}{\underbrace{\ 4x\ }} - \underset{\text{I}}{\underbrace{\ \ x\ \ }} - \underset{\text{L}}{\underbrace{\ \ 4\ \ }} = x^2 + 3x - 4 $

Then, after you've practiced a bit, you want to be able to combine the ‘outers’ and ‘inners’ in your head,
and write it down using only one step:

$(x-1)(x+4) = \underset{\text{F}}{\underbrace{\ x^2\ }} + \underset{\text{OI}}{\underbrace{\ 3x\ }} - \underset{\text{L}}{\underbrace{\ \ 4\ \ }} $

EXAMPLES:

Simplify: $(x+3)(x-2)$

Answer: x^2 + x - 6
Note: Key in exponents using the ‘ ^ ’ key.
Write your answer in the most conventional way.

Simplify: $(x+4)(x-4)$

Answer: x^2 - 16

Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
More Complicated FOIL