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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