Intro to polynomial synthetic division

In this expression, we're dividing this third degree polynomial by this first degree polynomial. And we could simplify this by using traditional algebraic long division. But what we're going to cover in this video is a slightly different technique, and we call it synthetic division. And synthetic division is going to seem like a little bit of voodoo in the context of this video. In the next few videos we're going to think about why it actually makes sense, why you actually get the same result as traditional algebraic long division. My personal tastes are not to like synthetic division because it is very, very, very algorithmic. I prefer to do traditional algebraic long division. But I think you'll see that this has some advantages. It can be faster. And it also uses a lot less space on your paper. So let's actually perform this synthetic division. Let's actually simplify this expression. Before we start, there's two important things to keep in mind. We're doing, kind of, the most basic form of synthetic division. And to do this most basic algorithm, this most basic process, you have to look for two things in this bottom expression. The first is that it has to be a polynomial of degree 1. So you have just an x here. You don't have an x squared, an x to the third, an x to the fourth or anything like that. The other thing is, is that the coefficient here is a 1. There are ways to do it if the coefficient was different, but then our synthetic division, we'll have to add a little bit of bells and whistles to it. So in general, what I'm going to show you now will work if you have something of the form x plus or minus something else. So with that said, let's actually perform the synthetic division. So the first thing I'm going to do is write all the coefficients for this polynomial that's in the numerator. So let's write all of them. So we have a 3. We have a 4, that's a positive 4. We have a negative 2. And a negative 1. And you'll see different people draw different types of signs here depending on how they're doing synthetic division. But this is the most traditional. And you want to leave some space right here for another row of numbers. So that's why I've gone all the way down here. And then we look at the denominator. And in particular, we're going to look whatever x plus or minus is, right over here. So we'll look at, right over here, we have a positive 4. Instead of writing a positive 4, we write the negative of that. So we write the negative, which would be negative 4. And now we are all set up, and we are ready to perform our synthetic division. And it's going to seem like voodoo. In future videos, we'll explain why this works. So first, this first coefficient, we literally just bring it straight down. And so you put the 3 there. Then you multiply what you have here times the negative 4. So you multiply it times the negative 4. 3 times negative 4 is negative 12. Then you add the 4 to the negative 12. 4 plus negative 12 is negative 8. Then you multiply negative 8 times the negative 4. I think you see the pattern. Negative 8 times negative 4 is positive 32. Now we add negative 2 plus positive 32. That gives us positive 30. Then you multiply the positive 30 times the negative 4. And that gives you negative 120. And then you add the negative 1 plus the negative 120. And you end up with a negative 121. Now the last thing you do is say, well, I have one term here. And in this plain, vanilla, simple version of synthetic division, we're only dealing, actually, when you have x plus or minus something. So you're only going to have one term there. So you separate out one term from the right, just like that. And we essentially have our answer, even though it seems like voodoo. So to simplify this, you get, and you could have a drum roll right over here, this right over here, it's going to be a constant term. You could think of it as a degree 0 term. This is going to be an x term. And this is going to be an x squared term. You can kind of just build up from here, saying this first one is going to be a constant. Then this is going to be an x term, then an x squared. If we had more you'd have an x to the third, an x to the fourth, so on and so forth. So this is going to be equal to 3x squared minus 8x plus 30. And this right over here you can view as the remainder, so minus 121 over the x plus 4. This didn't divide perfectly. So over the x plus 4. Another way you could have done it, you could have said, this is the remainder. So I'm going to have a negative 121 over x plus 4. And this is going to be plus 30 minus 8x plus 3x squared. So hopefully that makes some sense. I'll do another example in the next video. And then we'll think about why this actually works.