Using the FOIL Technique to Multiply Binomials

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 is 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 are taking something that seems to have two parts, and you are treating it as a single thing, a singleton!

Look what happens:

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 is 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:

\((x−1)(x+4)\)

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

\(=x^{2}+4x-x-4\)

\(=x^{2}+3x-4\)

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

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

Then, after you have 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:

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

Examples

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

Answer: \(x^{2}+x-6\)

Write your answer in the most conventional way.

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

Answer: \( x{^2}-16\)

Source: Tree of Math, https://www.onemathematicalcat.org/algebra_book/online_problems/foil_1x.htm
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