Dividing polynomials of degree one | Algebra 1 (TX TEKS) | Khan Academy

What we're going to do in this video is get some practice dividing expressions. So, what do I mean by that? So let's say that I have the expression 6X + 12, and I want to figure out what that divided by, maybe I'll write this in a different color: divided by 2x + 4 is. If you are so inspired, I encourage you to pause this video and try to work through this yourself before we do this together.

All right, so first of all, we can—there's many ways to write this same thing. We could have also written this as (6X + 12) divided by (2x + 4). When you divide something by something else, and let's say we get a question mark right over here, or we could say we do this division and we get a question mark over here.

Another way to think about it, let's imagine if we multiplied both sides by 2x + 4. Another way to think about an equivalent statement is that 6X + 12 is going to be equal to this question mark times (2x + 4). So, another way to think about it is: what can I multiply by (2x + 4) to get to (6X + 12)? Well, here you might say, “All right, to go from 2x to 6X, I would have to multiply by three.” So that's multiplying by three, and then to go from four to twelve, I'm also multiplying by three.

So, if my question mark is three, it feels like it's going to work. If I take three times (2x + 4), we can try it out. Well, we can see that when we distribute that 3, 3 * 2x is going to be equal to 6X, and then 3 * 4 is going to be equal to 12. So, the answer to our original question, what is (6X + 12) divided by (2x + 4)? Well, that's going to be three. Our question mark is three in this situation.

Now, the example that I just gave you—that's a situation where we don't have a remainder. But now, let me give you an example that does have a remainder. So, let's think about if I have—I'll try to get the colors again—(6A - 11) divided by, let's go with purple, that's always a little fun—divided by (2A - 5). Actually, I didn't have to write the parentheses the way I wrote it right over here. But one way to think about it, just like before, we could say that this is equal to question mark.

Or another way to think about it: if we multiplied both sides by (2A - 5), or you could say if I have question mark times (2A - 5), then I am going to get (6A - 11). That is going to be equal to (6A - 11). I don't have to write parentheses at this point.

So, we could do the same idea. We say, “All right, to go from 2A to 6A, I would multiply by three.” I'm multiplying by three. Now, we might be tempted to say, “Well, what happens if I multiply 5 by 3?” Well, that's not going to get us the exact same thing. That's actually going to be 15. Or, I have this negative sign out here—not one, and so this isn't going to be exact. But we can write it with a remainder.

Let's say that this is going to be roughly three. Let's figure out what we get when we multiply (3 * (2A - 5)), and then from that maybe we can figure out the remainder. So, if I take three times (2A - 5), that—and we just talked about it—is going to be equal to 3 * 2A is 6A, and then we could say, and then we're subtracting 3 * 5 which is 15. So, (6A - 15) that is obviously different than what we saw up here.

But what if we could rewrite what we have up here? So we have (6A - 15) and then plus or minus something else. So let's try to do that. So, I'm going to rewrite what we have up here. So, I have (6A - 11). Well, how could I write that with a -5 and then maybe a plus something else? Well, -11 is the same thing as -15 and then +4.

Remember, I'm just rewriting, looking for colors here. I'm just rewriting this (6A - 11) as the same thing as subtracting 15 and then adding 4. This is the same thing as -11 there. And if we write it that way, something very interesting happens because I could put the parentheses here. I could also say, “Well, let's just subtract the 15 first,” and then all of that, of course, is divided by (2A - 5).

Well, we could also rewrite this as (6A - 15) over (2A - 5) plus (4) over (2A - 5). All I did think about if I was going in the other direction, I have the same denominator here, so I could just add the numerators. Now, what we actually did in this situation is we started with the numerators being added and we say all this is the same thing as (6A - 15) over (2A - 5) plus (4) over (2A - 5).

But this is really useful because we just figured out that this whole thing right over here is the same thing as three. So we could say that our original question, what is (6A - 11) / (2A - 5)? We could say that this is the same thing as three plus (4) over (2A - 5). So you could view this right over here as the remainder. It goes three times with the remainder of this.

I know this is a little involved. You might have to watch this a couple of times to get exactly the intuition of what just happened. But in the future, we're going to find more systematic ways of doing this that maybe will require a little—it'll be a little bit easier to do—but it's really good to understand what we just did.