Common Core Algebra II.Unit 10.Lesson 10.Polynomial Long Division
Feb 22, 2017
Learning the Common Core Algebra II.Unit 10.Lesson 10.Polynomial Long Division by eMathInstructions
Hello, I'm Kirk weiler. And this is common core algebra two. By E math instruction. Today, we're going to be doing unit ten lesson number ten on polynomial long division. Now, that kind of sounds a little scary. Polynomial long division. But keep in mind that what we're trying to really do throughout this entire unit with rational expressions. Is that we're trying to make the link between what we do with integers, such as simplifying fractions, multiplying dividing, things like that. And what we do with polynomials. Polynomials are to integers as rational expressions aren't fractions. And one of the things that you learned how to do, granted, it was maybe a few years back, was how to divide two integers using long division. Today, we're going to be learning how to divide two polynomials by using exactly the same algorithm. So let's take a look. Okay, exercise one is all numerical. And in exercise one, it says consider the division problem 1519 excuse me. 1519 divided by 7, which could also be written as this, and it could be written this way as well.
Now letter aces find the result of this division using the standard long division algorithm. Is there a remainder in this division? So that's right, we are using the standard long division algorithm. So write this down. Okay, let's review how to do this. Now remember how you want to think about this. You want to say, well, the 7 go into one. The answer is no. How about 7 into 15? Sure. 7 goes into 15 two times. Now what do you do with that too? While you multiply it by 7, you get 14. Then what do you do? Subtract and you get one. Then what do you do? You have to drop this one down. And then you start all over. How many times does 7 go into 11? Well, it goes in there one time. So we get 7. Let me subtract and we get four. And then we bring the 9 down, and we get 9. We say how many times the 7 go into 49. And that's 7 times.
Let me do 7 times 7, and we get 49, and we get zero. And then we're done. Is there a remainder? No remainder. Or maybe it would be better to say a zero remainder. But most people would say, no remainder. So 1519 divided by 7 is 217. Now this next piece is very, very important. Letter B, which often is up here but down here on this worksheet says rewrite your result from letter a, so we're going to rewrite this result from letter a as an equivalent multiplication equation. All right? Now what does that really mean? It says, like somehow, I want to say 15 19 divided by 7 equals two 17. I want to write that equivalently in multiplication terms. How would you do that? Pause the video just for a second. All right. Well, what that really means is that 1519 is 7 times 217. That's it. Or likewise, 1519 is 217 times 7. Whichever way, all right.
Now that one was pretty easy because the remainder was zero. Now let's take a look at a division problem that has a remainder in it. All right, let's walk through it again together. Now we're going to do 1522 divided by 7. This is actually very close to what we did before. Except apparently I'm going to botch it. I'm going to have a two there and I'll have 14 and I'll subtract and I'll get one. And then I'll have 12 here and I'll put a one there and I'll put a 7 here and I'll subtract and I'll get 52. Now 7 doesn't go into 52 nicely, but the closest I can get is 7. And that's going to give me 49. And when I subtract, I'm going to have three. Now there's nothing left here to drop down, so I'm done. So what do I say? Well, I say that 15 22 divided by 7 is equal to 217. Do you remember this? With a remainder plus a remainder of three. Now, I've never really liked that that much, but think about what that really means. It means like if we had 1522 items. And we were putting them into 7 boxes. We'd have 217 in each box. 217 in each box. And then there would be three that would be left over, three that were remaining. But it's actually very, very important that you actually understand that 1522 divided by 7 is 217. Plus another three sevenths. Or if you want to write it and kind of mix numbers 217 and three sevenths. All right? I like this form the best. Okay?
Now, what does this mean in terms of what we did down here? I want to write down something very, very similar to this. Well, see if you can write down something similar. Right now, this fact 1522 divided by 7 equals 217 plus three sevenths. How can that be written as a multiplication equation? Well, here it is. It's 1522. Well, that's equal to 217 times 7. But plus an additional three. Right? Or 1522 is 7 times two 17. Plus an additional three. All right. Just a little review of numerical long division. Pretty soon, we're going to pop into polynomial long division. So pause the video now and make sure you understand this before we move on. Okay, I'm going to clear it out. All right, now let's see how it works when we divide two polynomials. Letter a. Simplify the following by performing polynomial loan division. All right, here it goes. Now we are going to do this as similar as we possibly can to what we were just doing. But of course, there's a big difference, because now you're looking at this and going, how do I even start? So let's talk about how you start. All right? All I want you to do is look at this X and look at this two X squared.
Whatever I put up here is going to then multiply the X plus 6, just like it did before. And I need to get rid of the two X squared. So I need to put something up here that when I multiply by X gives me two X squared. Let me say that again. Whatever I put up here, when it multiplies by X has to give me two X squared. And that's, of course, going to be a two X now remember, just like before, I'm going to multiply this. So two X times X is two X squared. Plus two X times 6 is 12 X and now I'm going to subtract. Now two X squared minus two X squared is zero. Actually, I'm not going to write that zero there. I'm going to now have 15 X -12 X and that's going to be three X I'm not adding them. I'm subtracting all of this. Now for some people that can be very tricky. So if that's a problem for you, what I would do is just change the signs and add. And then you add, okay? I'm going to go back to blue.
Then I drop this down. Plus 18. What do I do next? I now look at this three X and this X and what do I put up here? I put plus because I have to have it positive. Three. Why do I put plus three? Because three times X is three X, right? I got to match those up. These two have to match. Then three times 6 gives me 18. And now when I subtract three X minus three X is zero, 18 -18 is zero. So I get zero. So in other words, two X squared plus 15 X plus 18 divided by X plus 6 equals two X plus three. Right? But part B says write the result you found in a as an equivalent multiplication equation. You should be getting pretty good at this. Pause the video now and rewrite it. All right. Well, it just means the following. Two X squared plus 15 X plus 18. Equals two X plus three. Times X plus 6. Now what we really were just able to do is we just use this to factor this. And in fact, this will be the first time of many times at point this out.
The fact that there was a zero remainder. Means that what we were dividing by X plus 6 was a factor of two X squared plus 15 X plus 18. All right, that's what a factor is really are. Factors are numbers that when you divide other numbers, results in no remainder, zero remainders. All right? Now let's deal with the case. Where we do have a remainder. See how similar this is to the last problem. All right? It says write two X squared plus 15 X plus 20 divided by X plus 6. In the form, Q of X plus R divided by X plus 6, where Q of X is a polynomial. We should know what that means. R is a constant by performing long division. All right. So let's do it again. Plenty of practice and long division today. But I'm going to take this very, very slowly. Now again, it's going to play out very similar to this one, almost identical. So I need to figure out what I multiply X by to get two X squared. That's two X so I get two X squared. Plus 12 X, I subtract two X squared minus two X squared is zero, 15 X -12 X is three X again, I want to make sure that this matches up, so I have to put a plus three.
So three times X is three X three times 6 is 18. But now when I subtract three X minus three X is zero and 20 -18 is two. All right. That now means that two X squared, so this is my remainder. Right? That means now that two X squared. Two X squared plus 15 X plus 20 all divided by X plus 6, well, that's equal to two X plus three. That's my Q of X. Plus my remainder, which is two, divided by X plus 6. This is what in the business we call the quotient remainder form. The quotient remainder form of a rational expression, right? The quotient being this part, let me circle it in red. And the remainder, being this part. Okay. Now you're probably still a little bit confused about just how to do polynomial long division. So we're going to work with that a lot more. Don't worry about it. Backs out of the sheet, many more examples. And then we'll be doing quite a few in the next lesson as well. But take a look at this, pause the video and write anything down you need to. Okay, let's clear out the text. Backs out of the sheet. All right, exercise three.
Each of the following rational expressions in the form Q of X plus R divided by a form. So we're going to write these in their quotient remainder forms. Okay? So let's do it. All right. Yeah, I think we'll stick with black. That works just fine. Let's do X squared plus two X -5, and almost looks like a square root. I apologize for that. It's not meant to. Let me get rid of this little teeny tiny tail there. All right, let's do it. Now again, the way that you're going to do this is you want to get rid of the X squared. So I'm going to put an X here because X times X is X squared. Then I'm going to do X times negative three, and I'm going to get negative three X now this is very dangerous at this point. Because I'm subtracting a subtraction. All right? This is where most students find it way more helpful to change the sign on both of these and then add, all right? So when I add X squared and negative X squared cancel and then I get a 5 X here. I'm going to drop down that negative 5 now I have to think about the 5 X so I'm going to put a plus 5 here and there's going to be a 5 X -15 and again I have to subtract and that confuses students a lot of times.
So it's sometimes easier to just turn that into a negative, turn that into a positive get rid of this and add 5 X plus negative 5 X's gone and negative 5 plus 15 is positive ten. And that's it. That's my remainder. So at the end of the day, my final answer here is X plus 5 plus ten divided by X minus three. Now one thing I realized in the last problem was I never wrote that last one out as an equivalent multiplication problem. I apologize for that. It's very difficult for me to go back and correct that now. This is my final answer in this problem. In other words, X squared plus two X -5 divided by X minus three is equal to this. But if I wanted to write this as an equivalent multiplication statement, it would be the following. X squared plus two X -5 is X plus 5 times X minus three. Plus ten. Okay, this is the equivalent to this. In terms of a multiplication sentence. All right. I wanted to loop back from that last one. Lots of different colors. Why don't you try letter B on your own? Pause the video now. Okay, let's work through it. All right, we have two X squared -23 X plus 17.
I'm going to divide that by X minus ten. I had to get rid of the two X squared. I'm going to put a two X up here. Two X times X is two X squared. Two X times negative ten is negative 20 X again, subtracting a negative. If it's a problem for you, make this negative, make this positive. And add. Those two eliminate, going back to black, this will leave me with a negative three X plus 17. I've got a match up that negative three X because I'm going to be subtracting it from itself. So I'm going to get a negative three, which is going to give me negative three X plus 30. All right, I need to subtract. Again, it may be easier to change the signs. And then add. In that case, these cancel 17 plus negative 30 plus negative 30 is negative 13. There's my remainder. All right? So my final answer. Is going to be two X minus three -13 divided by X minus ten. Quotient remainder form. Two X minus three -13 divided by X minus ten. It's also completely okay to say plus negative 13 divided by X minus ten, either way. It doesn't really matter.
This may be this is even more more appropriate because then you can really see that your remainder is negative 13. All right, pause the video now and write down anything you need to. Okay, I'm going to clear out the text. Exercise four. Now, so far we've been doing polynomial long division in order to write things in quotient remainder form. Some of the problems you can do by just taking advantage of, let's say, some structure and some kind of kind of cool manipulation, all right? And let's take a look at that because sometimes that's easier than polynomial long division. You could still do polynomial long division on these, but sometimes it's easier to do it in a different way. Let's take a look at this very simple expression. X plus 8 divided by X plus three. It says write the numerator as an equivalent expression the numerator, this guy. I want to write an equivalent expression that involves X plus three.
Well, one way to do that is to take my 8 right and break it into three plus 5. I think everybody would agree that 8 is the same as three plus 5. But one of the great things and maybe I'll leave it like this. One of the great things about addition is it's associative. So I can make that X plus 8 into an X plus three plus 5. So let me just stop here. Now it says use the fact that division distributes over addition to write the final answer. In other words, I can take this thing. And I can say, well, I'll take X plus three and divide it by X plus three. And then I'll take 5 and divide it by X plus three. But X plus three divided by X plus three is one. And now I have it. There is. I don't know why I have this X plus 5 here, which should be a plus three. My apologies. But that's it. All right, simple enough. Take a look at what we have. And then we're going to move on to another one like this. All right. Never be afraid to break up addition into addition of smaller components. Okay?
Let's take a look at some that are a little bit more challenging. I like these a lot, okay? Now again, what I'd like to do is somehow work this out so I don't have to do polynomial long division. And somehow that X plus two is going to have to cancel. So ultimately, what I'd kind of like is something that might look like this. Four times X plus two plus something. All divided by X plus two. I'm kind of predicting where I want to go with this, right? Because then when I break it up, those X plus twos would actually cancel. Well, what is four times X plus two? Four X plus 8. Ah. So I could write that numerator. As four X plus 8 plus 5, right? Because these two added together are 13. And now I can break this up as four X plus 8. Plus 5, again, addition is associative. And now I can distribute that division. So I'd get four X plus 8 divided by X plus two. Plus 5 divided by X plus two. And if I factor this. Empty void, that was canceled. And I'll get four plus 5 divided by X plus two. Isn't that cool? We're taking advantage of the fact that addition is associative.
To break things up so that we can easily cancel factors. Easily write it in quotient remainder form. Now this only really you could do this every time instead of polynomial long division. Here it's relatively simple. Pause the video now and see if you can do letter B. All right, well, as usual, I gave you guys the hard one. Again, the idea is there's nothing much we can do about this three. Ultimately, I'd like somehow to have three times X minus four in the numerator. And that's three X -12. So that's what I want. That's what I'd like to get. All right, so let's play around with that. Well, if I had a three X -12 in the numerator, what else would I have to have so that I really had a -5? Well, that would be a plus 7, right? So negative 12 and positive 7 is negative 5. And then I'd have the X minus four in the denominator. And again, all I'm really doing is taking negative 5 and breaking it into negative 12 and positive 7. Now again, some students might say, well, why? How do you know you should break it into negative 12 and positive 7? And that's because I've been doing this thinking up here. I know ultimately I want to factor a three out and have a three times X minus four. But that takes a little bit of pre thought on it.
But now I've got it, right? I've got three X -12. Divided by X minus four plus 7 divided by X minus four. I'm really grouping the addition like this. And then I'm distributing the division to this. And to that, but then I'll have three times X minus four, all divided by X minus four. Plus 7 divided by X minus four, these cancel. And I get three plus 7. Divide it by X minus four. And there's my final answer. And it still is the same way as it was on my worksheet. All right. Pause the video now and think about these a bit. All right. Before I erase these, I do really want to point out that you could completely do these with polynomial long division. In fact, now that you've had a chance to kind of look at that, let me just for a moment, erase some of this. All right? And show you that you could definitely do this one because I know that this one's fairly challenging to do that way. You could definitely do it this way. And you could say, well, all right, I'll put a three here right three times four will be three X -12. I then would have to subtract, again, we got that whole negative thing going on, so maybe I'd do it this and this and add these two would cancel.
I'd get 7, just 7, and that would be my remainder. So I'd get three plus 7 divided by X minus four. That's a completely okay way to do it. Many of you will probably look at that and go, wow, that was much easier. That's fine. But I want to show you both methods in case one type of manipulation seems easier than the other one for you. All right, pause the video one last time, think about this, then we'll wrap up the lesson. Okay, clearing it out. All right, there is a big focus in common core math, especially high school math, trying to make parallels between integers and polynomials. And then between fractions and rational expressions. And we've been doing that all along the simplifying, the adding the subtraction multiplying dividing. And today, we reviewed how to do long division with integers, and then we extended that to dividing polynomials by one another. We'll look in the next lesson at this one more time and see how it's used in what's called the remainder theorem. For now though, I'd like to thank you for joining me for another common core algebra two lesson by E math instruction. My name is Kirk weiler, and until next time, keep thinking and keep solving problems.