# 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 FOIL, we calculate that this is

4*x*^{2} – 8*x* + 8*x* – 16,

which simplifies to

4*x*^{2} – 16.

Hmm, sneaky. The stuff we got from multiplying the "first" and "last" terms stayed around, but the stuff from the "outer" and "inner" 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 FOIL, this is

*x*^{2} – *xy* + *xy* – *y*^{2}.

The product of the outer terms and the product of the inner terms are additive inverses, meaning that when you add them, you have zero. It is almost as if they have 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 (▲ + ■)(▲ – ■),

the product of the outer terms and the product of the inner terms will cancel out, and we'll be left with (▲)^{2} – (■)^{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 FOIL as usual.

### Sample Problem

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

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

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

We can simplify this a little to get

29*x*^{2} + 2 × 28*x* + 16.

In the above example, look at the terms we found from using FOIL: the product of the outer terms was the same as the product of the inner terms. 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* + *v*)^{2}.

This means we want to find (*x* + *y*)(*x* + *y*). Using FOIL, we find

*x*^{2} + *xy* + *xy* + *y*^{2},

which simplifies to

*x*^{2} + 2*xy* + *y*^{2}.

In general, whenever we're squaring a binomial of the form (▲ + ■),

the product of the outer terms will be ▲ × ■ and the product of the inner terms will also be ▲ × ■, so

(▲ + ■)^{2 }= (▲)^{2} + 2▲ × ■ + (■)^{2}

**Be careful:** (*x* + *y*)^{2} *is not the same* as *x*^{2} + *y*^{2}. We found that

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

There's a big, glaring "2*xy*" in there. Remember the Alamo, and remember the 2*xy*.

### 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 are 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 FOIL 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

(▲ – ■)

the product of the outer terms will be -▲ × ■, and the product of the inner terms will also be -▲ × ■, so

(▲ – ■)^{2 }= (▲)^{2} – 2▲ × ■ + (■)^{2}.