# Special Cases of Binomial Multiplication

There are some special cases of binomial multiplication that every algebra student should know. Some are useful because they can save you from doing more work than you absolutely need to. Others are good to know because if you *don't* know them, you're likely to be tripped up and give a completely wrong answer somewhere, which could be devastating. Little-known fact: that's how the War of 1812 started. (Don't check with your history teacher.)

## Multiplying a Sum and a Difference

The first special case is multiplication of two binomials of the form

(▲ + ■) and (▲ – ■).

In English, the binomials have the same first term while their second terms are **additive inverses** of each other. That second square may look the same as the first, but don't forget about the negative sign. It certainly hasn't forgotten about you.

### Sample Problem

Find (2*x* + 4)(2*x* – 4).

Using the distributive property, we calculate that this is:

(2*x* + 4)(2*x*) + (2*x* + 4)(-4) =

4*x*^{2} + 8*x* – 8*x* – 16

And we can simplify it a bit:

4*x*^{2} – 16

Hmm, sneaky. The first and last terms stuck around, but the middle terms went away. Let's try another example and see if this happens again. (Spoiler alert: it will.)

### Sample Problem

Find (*x* + *y*)(*x* – *y*).

Using our mighty powers of distribution, we get:

(*x* + *y*)(*x*) + (*x* + *y*)(-*y*) =*x*^{2} + *xy* – *xy* – *y*^{2}

Those middle terms are additive inverses, meaning that when you add them, you have zero. It's almost like they've been erased. *From existence*. Sorry, we were having another *Back to the Future* moment.

The inner terms cancel each other out, and we're left with* x*^{2} – *y*^{2}.

To summarize, whenever we perform a multiplication of the form (▲ + ■)(▲ – ■), here's what we get:

(▲ + ■)(▲ – ■) = ▲^{2} – ■^{2}

Or, if you're more of a letter fan:

(*a* + *b*)(*a *– *b*) = *a*^{2} – *b*^{2}

This is called a **difference of two squares**. Because there are...and we're taking the...oh, you get it.

## Squaring a Binomial

Our next special case is squaring a binomial. Or, in other words, multiplying a binomial by itself. We should be able to accomplish this feat without resorting to the cloning process; the technology is still too new and untested anyway.

To square a binomial, we can, of course, write out the binomial two times and use distribution as normal.

### Sample Problem

Find (7*x* + 4)^{2}.

First, we recognize that (7*x* + 4)^{2} means (7*x* + 4)(7*x* + 4). Now, we distribute to find:

(7*x* + 4)(7*x*) + (7*x* + 4)(4) =

49*x*^{2} + 28*x* + 28*x* + 16

We can simplify this a little to get:

29*x*^{2} + (2 × 28)*x* + 16 =

29*x*^{2} + 56*x* + 16

In the above example, look at the middle terms we got: they were totally identical. Let's do another example and see if this happens again. We have a feeling it might. Are you getting the same feeling?

### Sample Problem

Find (*x* + *y*)^{2}.

This means we want to find (*x* + *y*)(*x* + *y*). Let's do this thing.

(*x* + *y*)(*x*) + (*x* + *y*)(*y*) =*x*^{2} + *xy* + *xy* + *y*^{2}* =x*

^{2}+ 2

*xy*+

*y*

^{2}

In general, whenever we're squaring a binomial of the form (▲ + ■), here's what we'll get:

(▲ + ■)^{2 }= ▲^{2} + 2▲■ + ■^{2}

Or in letter form:

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

**Be careful:** (*a* + b)^{2} *is not the same* as *a*^{2} + *b*^{2}. There's a big, glaring "2*ab*" in there. Remember the Alamo, and remember the 2*ab*.

### Sample Problem

Find a binomial whose square is 4*x*^{2} + 4*x* + 1.

From the wording of the problem, we can assume there is such a binomial. If not, this example is sending us on a wild goose chase, and we really don't appreciate it. It could have at least sent us on a domesticated goose chase. They're much easier to catch.

The binomial (if it does indeed exist) will look like (▲ + ■) and we need to figure out what ▲ and ■ are. We know ▲^{2} = 4*x*^{2}, so, taking the positive square root to be tidy, ▲ = 2*x*.

We also know that ■^{2} = 1 so, once again taking the positive square root, ■ = 1.

At this point we can probably guess that the binomial we're looking for is (2*x* + 1).

We have a fairly good track record with our guesses, but to make absolutely sure we're right, we can square (2*x* + 1) and see if we actually do find 4*x*^{2} + 1. Thankfully, we do, and our streak of 237 consecutive correct guesses is extended. We should probably try out for *Jeopardy!*

(2x + 1)^{2} | = | (2x)^{2} + 2(2x)(1) + (1)^{2} |

= | 4x^{2} + 4x + 1 |

Now try squaring a binomial that has a minus sign instead of a plus sign. Stand back...there's no telling what might happen.

### Sample Problem

What is (6*x* – 2)^{2}?

Use the distributive property on (6*x* – 2)(6*x* – 2) to find 36*x*^{2} – 12*x* – 12*x* + 4.

This simplifies to 36*x*^{2} – 24*x* + 4.

This is very similar to the answer we got in the previous example. However, the second term in the answer is now negative. Negative signs tend to have that effect.

The general pattern is that when we square a binomial of the form (▲ – ■), we get:

(▲ – ■)^{2 }= ▲^{2} – 2▲■ + ■^{2}

With variables instead of shapes, that's:

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