Introduction to polynomial division

Earlier in your algebraic careers, you learned how to multiply polynomials. So, for example, if we had (x + 2) times (4x + 5), we learned that this is the same thing as really doing the distributive property twice.

You could multiply (x) times (4x) to get (4x^2). You could multiply (x) times (5) to get (5x). You could multiply (2) times (4x) to get (8x), and you could multiply (2) times (5) to get (10). If what I just did looks foreign to you, I encourage you to review multiplying binomials on Khan Academy.

This could be simplified as being (4x^2 + 13x + 10). What we will now do as we advance our algebraic careers is to think about how we go the other way around. How do we divide polynomials?

So, for example, we might see an expression like this: (4x^2 + 13x + 10) divided by (x + 2). Now, you might already know from your knowledge of multiplication and division that this is really reversing what we just did up here.

If (x + 2) times (4x + 5) is equal to this business, then this business divided by (x + 2) should be equal to (4x + 5). Even if you didn't have this information that we have up here, there are ways that you would have approached this.

One way is that you could have tried to factor what we have here in the numerator. You could have said, "Hey, let me maybe factor this by grouping." I must do that right over here: (4x^2 + 13x + 10). When you factor by grouping, you say, "Hey, can I think of two numbers whose product is equal to (4) times (10)?"

So I say (a \times b) is equal to (40), and whose sum is equal to (13). So (a + b) is equal to (13). Let's see what it could be. It could be (8) and (5), so (a) is equal to (8) and (b) is equal to (5).

This is just factoring by grouping right over here. Once again, if this is unfamiliar to you, I encourage you to review factoring by grouping on Khan Academy. So we can break apart this (13x) as an (8x) and a (5x), so we can rewrite this as (4x^2 + 8x + 5x + 10).

Notice it's just redoing what we did up here, but we're assuming that we don't even know about what we did up here. This someone just gave us this quadratic and said factor it. Here, you could say, "Alright, for these first two terms, I could factor out a (4x)."

So it becomes (4x) times (x + 2), and then these second two terms I could factor out a (5), so plus (5) times (x + 2). Then, I can factor out an (x + 2), so it becomes (x + 2).

I'll scroll down a little bit. It becomes (x + 2) times (4x + 5), which is exactly what we had up there. So you could rewrite this expression that involves divisions, sometimes called a rational expression, as we can rewrite this as (\frac{x + 2 \times (4x + 5)}{x + 2}).

As long as (x) does not equal (-2), we can divide the numerator and denominator by (x + 2), and we're going to be left with (4x + 5). We could constrain it; we could say (4x) not equaling (-2).

This is just a little bit of a primer. As we go deeper, we'll do many, many, many examples of this, and we'll also see that there are other techniques other than just factoring this numerator over here.

We're going to do it with higher degree polynomials, third degree polynomials, and we're going to learn something known as polynomial division. It's going to look an awful lot, and it's actually going to have a lot of similarities with the long division that you likely learned in probably fourth or fifth grade.

We're going to take expressions like (x + 2) and we're going to divide it into (4x^2 + 13x + 10). Instead of place value, we'll have our new notion of place value, which is around which degree term you're thinking about.

We're going to do things like, and we're going to do these completely in other videos: "Hey, how many times does (x) go into (4x^2)?" Hey, it goes (4x) times. So you'd write the (4x) there, and then you multiply (4x) times (x) to get (4x^2).

(4x) times (2) is (8x). Then you subtract these, and then you keep going just like you would typically do long division. (13x - 8x) is (5x), and then you bring that (10) down and then you say, "Hey, how many times does (x) go into (5x)?"

You say, "Hey, it goes five times." (5) times (x) is equal to (5x), (5) times (2) is equal to (10). Then you subtract this, and you're left with no remainder.

Notice you just saw when (x + 2) is divided into this, you get (4x + 5), just what we saw right over there. So we're going to explore these multiple techniques, including polynomial division, and we're also going to see what happens when you do have a remainder there.