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:
$(a+b)(c+d)$ | ||
$= \overset{z}{\overbrace{(a+b)}}(c+d)$ | (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) | |
$\cssId{s32}{= \underset{\text{F}}{\underbrace{\ ac\ }}} \cssId{s33}{+ \underset{\text{O}}{\underbrace{\ ad\ }}} \cssId{s34}{+ \underset{\text{I}}{\underbrace{\ bc\ }}} \cssId{s35}{+ \underset{\text{L}}{\underbrace{\ bd\ }}}$ |
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:
You want to be able to write this down without including the first step above:
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:
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.