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:

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

Source: Tree of Math, https://www.onemathematicalcat.org/algebra_book/online_problems/foil_1x.htm
This work is licensed under a Creative Commons Attribution-NonCommercial 2.5 License.

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

Answer: $$x^{2}+x-6$$

Write your answer in the most conventional way.

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

Answer: $$ x{^2}-16$$

Simplify

1. \((x+2)(x−7)\)

2. \((x+2)(x+7)\)

3. \((x+3)(x−2)\)

4. \((x−7)(x−8)\)

5. \((x+8)(x−6)\)

6. \((x+5)(x−10)\)

7. \((x+8)(x+10)\)

8. \((x+8)(x−10)\)

9. \((x+6)(x+7)\)

10. \((x−6)(x+8)\)
    

1. \(x^2−5x−14\)

2. \(x^2+9x+14\)

3. \(x^2+x−6\)

4. \(x^2−15x+56\)

5. \(x^2+2x−48\)

6. \(x^2−5x−50\)

7. \(x^2+18x+80\)

8. \(x^2−2x−80\)

9. \(x^2+13x+42\)

10. \(x^2+2x−48\)