# 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 *again* on the first term to distribute (12*x*^{2}):

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

Then use the distributive property *a whopping third time* on the last term to distribute (5*x*):

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

Now simplify everything:

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

And finally write it all down 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:

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. With a bit of practice and some elbow grease, you'll get quicker and hopefully cut down on the amount of pencil scribbling when you're rocking two binomials. Even if "elbow grease" does sound pretty disgusting now that we think about it.

Let's run through one more of these guys before we call it quits.

### Sample Problem

Find (4*x*^{2} + 3*x*)(5*x* – 6).

First we distribute (4*x*^{2} + 3*x*) over all the stuff in the second term:

(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. He'll feel so left out if we forget about him.

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

Now we clean things up a bit with one more round of simplifyin':

20*x*^{3} + 15*x*^{2} – 24*x*^{2} – 18*x*

Hey, look at that: we've got two *x*^{2} terms, so we can combine them to finish up:

20*x*^{3} – 9*x*^{2} – 18*x*

It's a thing of beauty, isn't it?