Polynomial division introduction | Algebra 2 | Khan Academy

We're already familiar with the idea of a polynomial, and we've spent some time adding polynomials, subtracting polynomials, multiplying polynomials, and factoring polynomials.

What we're going to think about in this video, and really start to think about, is the idea of polynomial division.

So, for example, if I had the polynomial—and this would be a quadratic polynomial—let's say (x^2 + 3x + 2), and I wanted to divide it by (x + 1), pause this video and think about what that would be. What would I have to multiply (x + 1) by to get (x^2 + 3x + 2)?

Well, one way to approach it is we could try to factor (x^2 + 3x + 2), and we've done that multiple times in our lives. We think about, well, what two numbers add up to three, and if I were to multiply them, I get two. The ones that might jump out at you are two and one.

So, we could express (x^2 + 3x + 2) as ((x + 2)(x + 1)), and then all of that is going to be over (x + 1).

And so, if you were to take ((x + 2)(x + 1)) and then divide that by (x + 1), what is that going to be? Well, you're just going to be left with (x + 2). This is going on, you have to put parentheses. This is going to be (x + 2).

If we want to be really mathematically precise, we would say, "Hey, this would be true as long as (x) does not equal (-1)" because if (x) equals (-1) in this expression or this expression, we're going to be dividing by zero, and we know that leads to all sorts of mathematical problems.

But as we see, for any other (x), as long as we're not dividing by zero here, this expression is going to be the same thing as (x + 2). This is because ((x + 2)(x + 1)) is equal to what we have in this numerator here.

Now, as we go deeper into polynomial division, we're going to approach things that aren't as easy to do just purely through factoring, and that's where we're going to have a technique called polynomial long division—sometimes known as algebraic long division.

If it sounds familiar, it's because you first learned about long division in fourth or fifth grade. It's a very similar process where you would take your (x + 1) and you would try to divide it into your (x^2 + 3x + 2).

I'm going to do a very quick example right over here, but we're going to do much more detailed examples in future videos. You look at the highest degree terms and say, "Okay, I have a first degree term and a second degree term here. How many times does (x) go into (x^2)?" Well, it goes (x) times, so you put the (x) in the first degree column.

Then you multiply your (x) times (x + 1). (x) times (x) is (x^2), (x) times (1) is (x), and then you subtract this from that. You might already start to see some parallels with the long division that you first learned in school many years ago.

When you do that, these cancel out. (3x - x), and we are left with (2x). Then you bring down that (2). So, (2x + 2), and they say, "How many times does (x) go into (2x)?" Well, it goes (2) times, so you have a plus two here.

Two times (x + 1): two times (x) is (2x), two times (1) is (2). You can subtract these, and then you are going to be left with nothing. (2 - 2) is (0), (2x - 2x) is (0).

In this situation, it divided cleanly, and we got (x + 2), which is exactly what we had over there.

Now, an interesting scenario that we're also going to approach in the next few videos is: what if things don't divide cleanly? For example, if I were to add (1) to (x^2 + 3x + 2), I would get (x^2 + 3x + 3).

And if I were to try to divide that by (x + 1), well, it's not going to divide cleanly anymore. You could do it in either approach. One way to think about it if we know we can factor (x^2 + 3x + 2): say, "Hey, this is the same thing as (x^2 + 3x + 2 + 1)", and then all of that is going to be over (x + 1).

Then you could say, "Hey, this is the same thing as (\frac{x^2 + 3x + 2}{x + 1} + \frac{1}{x + 1})." We already figured out that this expression on the left, as long as (x) does not equal (-1), this is going to be equal to (x + 2).

So, this is going to be equal to (x + 2), but then we have that one that we weren't able to divide (x + 1) into. So, we're just left with (\frac{1}{x + 1}), and we'll study that in a lot more detail in other videos.

What does this remainder mean, and how do we calculate it if we can't factor a part of what we have in the numerator? As we do our polynomial long division, we'll see that the remainder will show up at the end when we are done dividing. We'll see those examples in future videos.