# Special Binomial Products

*See what happens when we multiply some binomials ...*

## Binomial

A binomial is a polynomial with two terms

example of a binomial |

## Product

**Product** means the result we get after multiplying.

In Algebra **xy** means **x** multiplied by **y**

And **(a+b)(a−b)** means** (a+b)**** **multiplied by** (a−b)**. We use that a lot here!

## Special Binomial Products

So when we multiply binomials we get ... Binomial Products!

And we will look at **three special cases** of multiplying binomials ... so they are **Special Binomial Products**.

## 1. Multiplying a Binomial by Itself

What happens when we square a binomial (in other words, multiply it by itself) .. ?

(a+b)^{2} = (a+b)(a+b) = ... ?

The result:

(a+b)^{2} = a^{2} + 2ab + b^{2}

This illustration shows why it works:

## 2. Subtract Times Subtract

And what happens when we square a binomial with a **minus** inside?

(a−b)^{2} = (a−b)(a−b) = ... ?

The result:

(a−b)^{2} = a^{2} − 2ab + b^{2}

## 3. Add Times Subtract

And then there is one more special case ... what about (a+b) times (a−b) ?

(a+b)(a−b) = ... ?

The result:

(a+b)(a−b) = a^{2} − b^{2}

That was interesting! It ended up very simple.

And it is called the "**difference of two squares**" (the two squares are **a ^{2}** and

**b**).

^{2}This illustration shows why it works:

a ^{2} − b^{2} is equal to (a+b)(a−b) |

Note: (a−b) could be first and (a+b) second:

(a−b)(a+b) = a^{2} − b^{2}

## The Three Cases

Here are the three results we just got:

(a+b)^{2} |
= a^{2} + 2ab + b^{2} |
} | the "perfect square trinomials" |

(a−b)^{2} |
= a^{2} − 2ab + b^{2} |
||

(a+b)(a−b) | = a^{2} − b^{2} |
the "difference of squares" |

Remember those patterns, they will save you time and help you solve many algebra puzzles.

## Using Them

So far we have just used "a" and "b", but they could be anything.

### Example: (y+1)^{2}

We can use the (a+b)^{2} case where "a" is y, and "b" is 1:

(y+1)^{2} = (y)^{2} + 2(y)(1) + (1)^{2} = y^{2} + 2y + 1

### Example: (3x−4)^{2}

We can use the (a-b)^{2} case where "a" is 3x, and "b" is 4:

(3x−4)^{2} = (3x)^{2} − 2(3x)(4) + (4)^{2} = 9x^{2} − 24x + 16

### Example: (4y+2)(4y−2)

We know the result is the difference of two squares, because:

(a+b)(a−b) = a^{2} − b^{2}

so:

(4y+2)(4y−2) = (4y)^{2} − (2)^{2} = 16y^{2} − 4

Sometimes we can see the pattern of the answer:

### Example: which binomials multiply to get 4x^{2} − 9

Hmmm... is that the difference of two squares?

Yes!

**4x ^{2}** is

**(2x)**, and

^{2}**9**is

**(3)**, so we have:

^{2}4x^{2} − 9 = (2x)^{2} − (3)^{2}

And that can be produced by the difference of squares formula:

(a+b)(a−b) = a^{2} − b^{2}

Like this ("a" is 2x, and "b" is 3):

(2x+3)(2x−3) = (2x)^{2} v (3)^{2} = 4x^{2} − 9

So the answer is that we can multiply **(2x+3)** and **(2x−3)** to get **4x ^{2} − 9**