# Multiplication of Two Binomials

Multiplying two binomials is still an application of the distributive property. In fact, we can use the distributive property even more than we did in the first example. It's like the stuffing and mac 'n cheese that you go back for over and over again at Thanksgiving dinner, while the poor broccoli casserole languishes in the corner.

### Sample Problem

What is (4*x*^{2} + 3)(12*x*^{2} + 5*x*)?

We use the distributive property to distribute the expression (4*x*^{2} + 3) over the polynomial (12*x*^{2} + 5*x*), like this:

(4*x*^{2} + 3)(12*x*^{2}) + (4*x*^{2} + 3)(5*x*).

Now we have a new expression with two terms. We use the distributive property on the first term to distribute (12*x*^{2}) and find

(4*x*^{2})(12*x*^{2}) + (3)(12*x*^{2}) + (4*x*^{2} + 3)(5*x*),

then use the distributive property on the last term to distribute (5*x*) and get

(4*x*^{2})(12*x*^{2}) + (3)(12*x*^{2}) + (4*x*^{2})(5*x*) + (3)(5*x*).

Now simplify the terms to find

48*x*^{4} + 36*x*^{2} + 20*x*^{3} + 15*x*

and finally write in order of descending exponents:

48*x*^{4} + 20*x*^{3} + 36*x*^{2} + 15*x*.

By this point, you're probably stuffed. However, those au gratin potatoes certainly do seem to be calling your name...

### Sample Problem

What is (3*x* – 1)(4*x* + 2)?

Distribute (3*x* – 1) over (4*x* + 2):

(3*x* – 1)(4*x*) + (3*x* – 1)(2).

Now distribute (4*x*) over (3*x* – 1) and distribute (2) over (3*x* – 1):

(3*x*)(4*x*) – (1)(4*x*) + (3*x*)(2) – (1)(2).

Simplify the terms for

12*x*^{2} – 4*x* + 6*x* – 2

and combine like terms to get

12*x*^{2} + 2*x* – 2.

The terms are already written in descending order by exponent, so we're done. Delicious...time for dessert!

Clearly, there's a whole lotta distributin' going on. Thankfully, there's a shortcut for multiplying two binomials that lets us do all the distributing at once and cuts our pencil scribbling in half. Think of it as Weight Watchers for binomials.

To remind ourselves what we won't be missing, we'll do one more example the long way. Hate it while it lasts.

### Sample Problem

Find (4*x*^{2} + 3*x*)(5*x* – 6).

Distribute (4*x*^{2} + 3*x*) to find

(4*x*^{2} + 3*x*)(5*x*) – (4*x*^{2} + 3*x*)(6),

then distribute (5*x*) and (6), remembering to distribute the minus sign as well, to get

(4*x*^{2})(5*x*) + (3*x*)(5*x*) – (4*x*^{2})(6) – (3*x*)(6).

Now, before simplifying, look at what we wound up with after all those uses of the distributive property. Each term from the first polynomial got multiplied by each term from the second polynomial. Hm...

This leads us to our shortcut, commonly known as **FOIL**. Rather than foiling our plans, this helpful acronym will actually make things much easier on us.

FOIL stands for "First, Outer, Inner, Last," and reminds us how to combine the terms of two binomials in order to multiply them. (Note: FOIL should never be performed anywhere near a microwave. Trust us on this one.)

The words "First, Outer, Inner, Last" are mostly self-explanatory, but since we like demonstrating things, we'll take the product of binomials (x + 4)(2*x* - 3) as an example.

The *first* terms are the first terms in each binomial: (* x* + 4)(

**2**- 3).

*x*The *outer* terms: (* x* + 4)(2

*x*

**- 3**).

The *inner* terms: (*x* + **4**)(**2 x** - 3).

And, finally, the *last* terms: (*x* + **4**)(2*x*** - 3**).

### Sample Problem

Find (2*x* – 7)(3*x* + 4).

We'll use FOIL. We just learned it, so it seems like a waste not to.

**F:** We multiply the *first* terms of each polynomial for (2*x*)(3*x*) = 6*x*^{2}.

**O:** We multiply the *outer* terms for (2*x*)(4) = 8*x.*

**I:** We multiply the *inner* terms for (-7)(3*x*) = -21*x.*

**L:** Finally, we multiply the *last* terms of each polynomial for (-7)(4) = -28.

Adding the results of all the little multiplications yields

6*x*^{2} + 8*x* – 21*x* – 28,

which simplifies to

6*x*^{2} – 13*x* – 28.

That was much easier. We owe one to FOIL. Or, in acronym form, WOOF.