by Dr. Carol JVF Burns (website creator)
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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 ‘treat it as a singleton’ 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:

      $= \overset{z}{\overbrace{(a+b)}}(c+d)$ (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)
  $\cssId{s32}{= \underset{\text{F}}{\underbrace{\ ac\ }}} \cssId{s33}{+ \underset{\text{O}}{\underbrace{\ ad\ }}} \cssId{s34}{+ \underset{\text{I}}{\underbrace{\ bc\ }}} \cssId{s35}{+ \underset{\text{L}}{\underbrace{\ bd\ }}}$  

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:

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:

      $\cssId{s62}{= \underset{\text{F}}{\underbrace{(x\cdot x)}}} \cssId{s63}{+ \underset{\text{O}}{\underbrace{(x\cdot 4)}}} \cssId{s64}{+ \underset{\text{I}}{\underbrace{(-1\cdot x)}}} \cssId{s65}{+ \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:

$\cssId{s69}{(x-1)(x+4)} \cssId{s70}{= \underset{\text{F}}{\underbrace{\ x^2\ }}} \cssId{s71}{+ \underset{\text{O}}{\underbrace{\ 4x\ }}} \cssId{s72}{- \underset{\text{I}}{\underbrace{\ \ x\ \ }}} \cssId{s73}{- \underset{\text{L}}{\underbrace{\ \ 4\ \ }}} \cssId{s74}{= 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:

$\cssId{s78}{(x-1)(x+4)} \cssId{s79}{= \underset{\text{F}}{\underbrace{\ x^2\ }}} \cssId{s80}{+ \underset{\text{OI}}{\underbrace{\ 3x\ }}} \cssId{s81}{- \underset{\text{L}}{\underbrace{\ \ 4\ \ }}} $

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.