# 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}

If you want to see why, then look at how the (a−b)^{2} square is equal to the big a^{2} square minus the other rectangles:

(a−b)^{2} = a^{2} − 2b(a−b) − b^{2}

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

Note: (a−b) can swap places with (a+b):

(a−b)(a+b) = a^{2} − b^{2}

## The Three Cases

Here are the three results we just got:

^{2}= a

^{2}+ 2ab + b

^{2}

^{2}= a

^{2}− 2ab + b

^{2}

^{2}− b

^{2}

The first two are the "perfect square trinomials" and the last is 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} − (3)^{2} = 4x^{2} − 9

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