BASIC FOIL

Recall the the distributive law:   for all real numbers [beautiful math coming... please be patient] $\,a\,$, $\,b\,$, and $\,c\,$, [beautiful math coming... please be patient] $\,a(b+c) = ab + ac\,$.

At first glance, it might not look like the distributive law applies to the expression [beautiful math coming... please be patient] $\,(a+b)(c+d)\,$.
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:   [beautiful math coming... please be patient] $\,z(c+d) = zc + zd\,$.
This says that anything times [beautiful math coming... please be patient] $\,(c+d)\,$ is the anything times $\,c\,$, plus the anything times $\,d\,$.

Now, look back at [beautiful math coming... please be patient] $\,(a+b)(c+d)\,$, and take the group [beautiful math coming... please be patient] $\,(a+b)\,$ 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:

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

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 [beautiful math coming... please be patient] $\,(a+b)(c+d)\,$.

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

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

[beautiful math coming... please be patient] $\,(x-1)(x+4)$
      $= \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)}} $
      [beautiful math coming... please be patient] $= x^2 + 4x - x - 4$
      [beautiful math coming... please be patient] $= x^2 + 3x - 4\,$

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

[beautiful math coming... please be patient] $(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:

[beautiful math coming... please be patient] $(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

 
 

Answers must be written in the most conventional way:
$\,x^2\,$ term first, $\,x\,$ term next, constant term last.

Note:   Key in exponents using the ‘ ^ ’ key.

Simplify: