Understanding Long Division and Polynomial Division

We ended off last time with talking about long division, which has probably been a while for many of you. If you guys came in late, just please take a seat quickly and quietly, because I've already started. Okay? Um, shh. Okay, and the thing that I really wanted to sort of remind you of is firstly the process of taking a number, dividing it by a number, how we do that with long division, and then also the remainder. Okay, so, you know... We have up to, I'm getting really distracted when people are coming in late and then catching up with, like, I get it. But if you come in five minutes late, there were people here who were on time, and that's not fair. So, you come in late, not a problem. Just sit down and be quiet. Thank you. Okay. Okay, so... We looked at this one example. I can't remember what the number was. It was divided by four. But, you know, once you get beyond, I don't know, like grade four or five, you stop really thinking about, you know, let's say I did 22 over three. We don't really talk about it in terms of division. We think of it more in terms of a fraction, which is division. So what we would do with that, you know, from most of the time anyway, I don't know. from maybe, I don't know, grade five or six on, is we'd maybe convert it into an improper fraction. We may convert that into a decimal. It'd be 7.3333333. But I just kind of want to go back to grade three, four, whatever. Okay. And what you would have looked at before you knew what an improper versus a proper fraction is, before you knew what a repeating decimal was, you may not even know. known what a decimal was at that point. Okay? And what we would have done is said, okay, well, if I'm doing 22 divided by 3, how many times does 3 go into 22? How many whole times does 3 go into 22? How many whole groups of 3 can you make if you have 22? Seven. Seven. And then what's left over? One third, yeah. Right? And essentially, this is kind of the relationship between an improper fraction and a mixed number. Okay? Now, when we're doing it with something like this, that's a much bigger number. to work with. And so, yeah, we may have to do long division to be able to get to that mixed number stage. Okay. But this is kind of essentially where we all started. Okay. Now, if I take this, equation and let's say I multiplied the entire thing by 3, then another way of writing the relationship between all of these numbers, so I multiply this side by 3 I have 22, if I multiply this by 3 I have 21. I'm actually going to write that as 7 times 3. Plus, if I multiply that by 3, I have 1. And this is true. If I have 22, I've got 7 groups of 3 plus an additional 1. So now what I want to do is actually introduce some vocabulary around all of this stuff. When I take 22 and I divide it by 3, the number that I have originally, this 22, is called the dividend. Okay, the three is the divisor. Okay, the seven, what do you think that's called? It's called the quotients. And then this one. is the remainder. Okay. Now, when we start to do polynomial division, we're going to see a similar format, either in this first line or possibly like this as well. Okay, so let's talk about that. Next. Okay. So we've done a review of long division already. Polynomial division, just to put the last example, 22 over 3, with all of that vocabulary into context right now. What we're going to be doing here is taking some polynomial expression, and I'm going to call that p of x. And I'm going to divide it by another polynomial. That's the divisor. I'm going to call that d of x. Okay. Now what I end up with is some quotient, q of x, plus the remainder over the divisor. R of X over D of X. And most, almost always, that remainder is actually a constant. It's a numerical value, okay? So just so that you can kind of see where this all came from, it's like if I put an example in here, 22 over 3 equals 7 plus 1 third. Except instead of having numbers, these are going to be... Now, you may sort of wonder where we're going with this. Eventually, at the end of the day, what we'd like to be able to do is take a polynomial expression that's larger than degree 2, right? Because you guys are really good with quadratic expressions, ax squared plus bx plus c, and be able to factor it. And so we need a way to sort of start to pick apart these expressions by breaking them down into smaller parts so that we can have an expression that's completely factored, right? So think like even take the number 20 and breaking it down into prime factors. You'd go 2 times 10, you could break 10 down into 2 times 5. We're gonna be doing the same thing except, you know, it's just kind of like factoring quadratics or trinomials except we're gonna be starting with a polynomial of a higher degree. OK? And ultimately, what that will allow us to do is that if we have a function of a polynomial with higher degree, if it can be factored, we can find the roots from the factored form. OK? All right, so polynomial division, long division, works very, very similarly actually to numerical long division. I'm going to attempt to write out the steps, but know that it'll make way more sense when you actually see an example. So I actually don't think I'm going to write out every single step. I'm just going to write the beginning part. And the beginning part is essentially we... You set it up the same way as long division, with that sort of... half a box thing. And what we're going to do is divide the leading term of the dividend by the leading term of the divisor. Okay, so I don't know, let's say we're working with something that's a cubic polynomial, and this is like, the leading term is 3x cubed, and then we divide it by x plus 2. We're just going to take the leading term, the 3x, 3x cubed divided by x because that's the leading term of the divisor. Okay. Once you do that, you multiply through, subtract, bring down the next one, repeat. Okay. So that's the part where it's really hard to put it into step by step without an example beside it. So we're going to now multiply through. Subtract, bring down the next term, and then add heaps. Okay, so I know that that doesn't make a lot of sense right now, but hopefully when you see an example you'd be like oh yeah okay I kind of see what's going on here. Okay now one thing that is very important is that when you start with the dividend within the long division you have to see every power of x. So for example if you have a cubic polynomial you have to have an x cubed term and x squared term, an x term, and a constant. And if you are missing any of those in the original, you're going to put in a placeholder. Like let's say there's no x squared term. You're going to add in zero x squared. So it's a placeholder. Okay. It's kind of similar to place value with numbers, right? If we have the number 3,021, we write that as 3,021 to show that there are 0 hundreds in that number. It's a similar sort of idea. Okay? Okay, let's give this a try. Okay, so here is our polynomial, our dividend 2x cubed plus 3x squared minus 4. and we are going to divide that by the divisor x minus 1. And ultimately, the goal is to write it in this form. Okay? So here's how we're going to start. We're going to set up a nice long division symbol. This is what I'm dividing by, so this is going to go over here, x minus 1. And then I have to put this polynomial under the division sign. Now this is cubic. I have an x cubed term. I have an x squared term. I don't have an x term. So I'm going to add in 0x. as a placeholder. And then I have the minus 4. So this is going to be 2x cubed plus 3x squared plus 0x minus 4. Yes? That's what we're talking about, yeah. Okay? Alright, so first step, we take the leading term of the dividend. That would be the 2x cubed and we're gonna divide it by the leading term of the divisor, which is x. Okay, so what is 2x cubed divided by x? 2x squared. So we're gonna put that up here. Okay, now we multiply through. This times that goes there. So 2x squared times x is 2x cubed. 2x squared times negative 1 is minus 2x squared. Yes? How did you get 2x squared? I went 2x cubed divided by x. Yeah, leading term of this divided by leading term of that. Okay? All right, so I multiply through. I'm going to subtract. This by design is 0. That's what I want. But 3x squared minus negative 2x squared is 5x squared. Okay, now bring this down. Plus 0x. Okay, so now I'm going to repeat. And I'm going to go, okay, 5x squared divided by x. What's 5x squared divided by x? 5x. So I'm going to add 5x. I'm going to multiply through. 5x times x is 5x squared. 5x times negative 1 is minus 5x. And I'm going to subtract. Okay. Now again, by design, this is 0. 0x minus negative 5x is 5x. Yep. This one? So I put a 5x over there, and now I'm multiplying through 5x times x minus 1. That goes there. Yeah. Okay. All right. So I subtracted. I'm left with 5x. Bring this down. Minus 4. And I'm going to do the same thing. Okay. So now I'm dividing the leading term here as 5x, dividing that by x. 5x. Divide by x, 5. And I'm going to multiply through. 5 times x minus 1 is 5x minus 5. And when I subtract, this is 0. This is negative 4 minus negative 5 is 1. That's my remainder. Okay, so this over here is the dividend. This, divisor. This over here, quotient. That's the remainder. Okay, now in this format, okay, so p of x was my original polynomial. Now at this stage, you can take out the 0x. If you want to put it in, that's fine, but you don't have to have it in there. Okay, so that was 2x cubed plus 3x squared. minus 4, oh I didn't give myself very much room, I'll have to do it on the next line, over the divisor is x minus 1, equals the quotient 2x squared plus 5x plus 5, plus the remainder 1 over x minus 1. Okay, now I know that this can seem abstract, but if you remind yourself how it works with numbers, if we were to do the same thing with like 22 divided by 3, right, we would do this. Or 22, sorry. This goes in 7 times. 7 times 3 is 21. Remainder 1. So 22 over 3 equals 7 plus 1 over 3. Okay. Yes? Is there any, like, practice sheet? Is there any white? There's a lot of practice, yes. That's going to be your homework today. Yes, okay. So, that's polynomial long division. Okay? Now, in reality, we have some shortcuts that we can take. And essentially, like to understand why the shortcut works, you actually have to go back to what I talked about Last class in that in fact place value is a condensed way of writing a number based on powers of 10 and So when we actually write the number out. We don't write this whole thing. We write 3,241. 3, 4, or 3, 2, 4, 1. Okay. Now, when we did the polynomial long division, we didn't do that. We kept all those power of x in there. Synthetic division essentially removes all of the powers of x and just keeps the coefficients. Just like when we're working with the number 3,241, we don't write it as 3 times 10 cubed plus 2 times 10 squared, et cetera. All right, now, synthetic division. Same process, except we are taking a few shortcuts so that we don't have to write all of the powers of x down. Okay, so here's how it works. What you're going to do is set up an upside down division sign. And over here in the top line, there should be enough room for sort of two lines. In the top line, you are going to put just the coefficients of the dividend in descending order. So I'm going to put in 2, 3, 0, negative 4. Okay, what I'm going to put here is if this is written in the form x minus a, I'm going to write just the negative one piece. Okay. So now, here's what we're going to do. Bring this down. Multiply. 2 times negative 1, negative 2. Okay? Now, subtract. 3 minus negative 2 is 5. Multiply. 5 times negative 1 is negative 5. 0 minus negative 5 is 5. Multiply again. 5 times negative 1 is negative 5. Subtract. 4 minus negative 5 is 1. Okay, so now we're done. Now here's how we interpret what we ended up with. This is the remainder. These three numbers, 2, 5, 5, make up the coefficients of the quotient. Okay, so it was 2x squared plus 5x plus 5. Remainder 1. Okay, so again... 2x cubed plus 2x, sorry, 3x squared minus 4 is equal to, okay, that's the quotient. 2x squared, sorry, over x minus 1. 2x squared plus 5x plus 5. And then the 1 is the remainder. Plus 1 over x minus 1. Okay. Now, I do want to warn you, if any of you guys have tutors or look at YouTube videos or whatever, there are two ways to do synthetic division. Well, it's all pretty much the same, but here's how things differ. Some people keep this sign. and maintain it. If you do that, you subtract. Some people change that to a positive one, and then instead of subtracting, they add. Okay? You end up with the same quotient and remainder at the end of the day. Okay? I was actually taught to change the sign and add, but I kind of like the idea of not changing the sign and subtracting because it It models the long division process more closely. But you guys can do whatever you want. Yes? If it was x plus 1, would you add? No, you always subtract. The question is, do you change this side? sign from the x minus a? Do you keep that negative one or do you change it to a one? Okay. Now I'm going to stick with this where I don't change the sign. I subtract because again, in long division, that's what you do. You subtract. But I just want you guys to know that I would say it's a pretty much 50-50 split between who does it change in the sign and adding and who keeps the sign and subtracting because otherwise you guys are going to come to me and be like, wait, how come these people are subtracting? are adding instead of subtracting, okay? And you can do it either way. Yeah? I know we just did that question. Yes. Yes. Yes. How do you know it's 2x squared? Okay, good question. So what you want to keep in mind is that in the first step of long division, you're dividing this leading coefficient by this guy. So this is always going to be one. degree less than whatever that is. Okay, all right, so let's try this. Now you guys are sort of expected to know both processes, writing out the full polynomial long division and also just writing out synthetic division. Ultimately though most people just do synthetic division because it's a lot faster. Yes? Yes. Okay. Now again, where this is leading, like we're not just doing this for fun. Ultimately, where this is going is here. Let's say we get a remainder of zero. What does that mean? If I did 21 divided by 3 instead of 22 divided by 3, what's my remainder? 21 divided by 3? Zero. Okay, so what does that tell me about the 3 and the 21? Yeah, it's a factor, exactly. So if we end up with a remainder of 0, what we know is that, OK, we have a factor. And so that's going to help us in the factoring process. Okay, now we still have lots to learn because all, you know, if you're just trying to see is something a factor, I mean, you kind of want to have a starting place and know, okay, well, what should I be testing out here? We'll get to that. But right now we're just learning this process. Okay. All right. Let's try this guy. Okay, so we have x cubed plus 7x squared minus 3x plus 4 divided by x minus 2. So I am going to set up this upside down division sign. And over here, I'm going to put that negative 2 value. Okay. So within here on the top line, I'm going to put the coefficients of my evident. And I'm going to make sure that I'm not missing any powers of x. Because if I'm missing a powers of x, I have to put a zero in as a placeholder. But it looks like I've got everything there. So I've got 1, 7, negative 3, and 4. And I'm going to start by bringing this down. Okay? And what we do now is multiply, subtract. Multiply, subtract. Multiply, subtract. Yes? A double, like if this was 14. Oh, if the coefficient was double digits, you just write them both down. Yeah. So in that way, I guess like it's not the same as place value because with place value. if you had 14 groups of 10 let's say you wouldn't write it that way you would one group of 100 and four groups of 10 that doesn't carry through there are some holes in the modeling so 1 times negative 2 is negative 2 subtract what do we get? 9 9 9 times negative 2, negative 18. Okay, subtract. 15. 15 times negative 2, negative 30. Subtract. Yes. For some reason in my head I had 26. 34. Okay. So in terms of, I mean, this is just a, like a graphic organizer to get us to our answer. This is not the answer though. Okay. Ultimately, the answer is... When we take this original polynomial, minus 3x plus 4, and divide it by x minus 2, we get a quotient of, this is 1x squared plus 9x plus 15. And then here's our remainder, plus 34 over x minus 2. Can you just do synthetic division? Okay, so the short answer is for this course, you need to be able to show both. In the long run, does anyone really do the full long division? No, because this is an efficiency, right? Yeah. Okay. So one other thing that I just want to show you is that this is one format that you could be asked for. Another way that you might see this being written is where you essentially Multiply the entire equation by the x minus 2 so that there's no denominators. And that was kind of like when I did, did I do it here? No, but I did 22 equals 7 times 3 plus 1. So how did I get that? I multiplied the whole thing by 3. So you could see that too. If I multiply the whole thing by x minus 2, this denominator is going to be gone. False. I'll have x cubed plus 7x squared minus 3x plus 4. This is the only one that doesn't have a denominator of x minus 2. So that's where I'm going to see the x minus 2 times this quotient. And then here I have this remainder of 34. What's that? Oh, okay, so what we were solving was a rational equation. It's a similar idea in that you're multiplying through by the least common denominator to isolate x. except that we're not solving for a value of x, really, that satisfies the equation. We're more manipulating the algebra, so it looks a different way. Yeah, but it's a similar idea in terms of multiplying through by the least common denominator. Okay? All right. Let's try this one. Okay, so where do I need to start? Yeah, sure. Let's do the long way. We can do both. Okay, so I have to rewrite it, though, for the long way. It's 3x squared. No, 3x cubed. And what is it? Plus 8x? Minus 8x. Minus 1? Okay. Over x plus 2. Okay? All right, so the long way... is I have to write it like division, right? So this is x plus 2. And then, but again, just like synthetic division, I still need a placeholder for the x squared term. So I'll write this as 3x cubed plus 0x squared minus 8x minus 1. Okay, and then, so what do I do first? Yeah, so take leading term divided by leading term. Okay, so 3x squared divided by x is 3x cubed, thank you, is 3x squared. Then I'm going to multiply through. So 3x squared times x plus 2 is 3x cubed plus 6x squared. And then I subtract. Okay, so this piece is 0. This is negative 6x squared. Then I bring this down, minus 8x. Okay, so what do I do next? Yeah? Almost. What do I need to multiply through, though, by? Yeah, I need to divide. Yeah, so I'm going to go 6x, negative 6x squared divided by x. And that's minus 6x. Okay, and now I can multiply through. So negative 6x times x is negative 6x squared. And negative 6x times 2 is minus 12x. Okay, and now I can subtract. Watch out for the double negatives, though, when you subtract, right? That's one place where sometimes people make careless mistake. Okay, so... Negative 8x minus negative 12x is 4x minus 1. And then the last piece, 4x divided by x is 4. And then 4 times x plus 2 is 4x plus 8. And we end up with... Negative 9. So that's our remainder. Now this is also different than how it would work with numbers because when we're working with numbers, our remainders will only be positive. Whereas in this case, what we're actually looking for is a constant. Okay? So. At the end of the day, how would we show the relationship between, there's four things, right? There's the dividend, the quotient, sorry, the divisor, quotient, remainder. And if you ever forget... where everything goes, do the 22 over 3 equals 7 plus 1 third. And that will help remind you. Okay? All right. So we're going to have 3x cubed minus 8x minus 1. over x plus 2 equals the quotient, which is 3x squared minus 6x plus 4 plus the remainder. So minus 9 over x plus 2. Okay, now just so that you can see the synthetic division and sort of follow it through this problem, if we were to do this with synthetic division instead, then I would still have those same coefficients, okay? This would be a 2. I'm bringing this down. Multiplying, 3 times 2 is 6. Subtracting, negative 6. Okay. Multiplying, negative 6 times 2 is negative 12. Subtracting, 4. 4 times 2 is 8. Subtracting is, oh, sorry, this is a minus 1. Negative 9. Okay. So. We kind of see like, okay, and then we have this and that. I'm trying to like color code it, but this and that, this and that. Okay, and then we have. The negative 12 and 4, and then the 4 and the 1, the numbers are all the same. We're just kind of... We're dropping the powers of x so that we don't have to keep track of those as well. Okay. All right. So. Here's an example. The volume of a rectangular prism is given by v of x equals x cubed plus 3x squared minus 36x plus 32. Determine possible measures for the width and the height in terms of x if the length is x minus 4. Okay. So... Let's just remind ourselves this is a rectangular prism. How do we get the volume of a rectangular prism? Yeah. Oh, I didn't do that. Volume is length times width times height. So what we are after is what could these two be? What could the width be and what could the height be? If we know the volume and the length, let's start out by just, I mean, what would W times H be? Let's just start with that. This part. What if I isolated the width times the height? What would you have to do to get- this piece by itself. Right now it's being multiplied by length, divided by length. So if I divide both sides by length, then whatever I get is the width times the height. So an equivalent type of question with numbers would be, okay, you have a rectangular prism, the volume... is 30. And the side lengths are all whole numbers. Okay? So now if I tell you the length is 5, what could the other two sides be? The volume is 30, the length is 5. 3 and 2, right? Because the width times the height would have to be 6. 5 times width times height is 30. 5 times 6 is 30. So the width times the height would have to be 6. to be 6. So it could be 3 and 2. It could also be 1 and 6. Similar type of question. All right. Well, let's start with just trying to get an expression for width times height. Width times height is going to be the volume x cubed plus 3x squared minus 36x plus 32 over x minus 4. OK, now, x minus 4 should be a factor, because we're told that it's the length. And the length should be a factor of the volume. So ultimately, if we divide these, we're dividing a cubic polynomial by a linear. polynomial, we should end up with something quadratic. If we have something quadratic, the hope is that we can factor it to get the other two sidelines. So we're starting with cubic, breaking it down into a binomial times a quadratic, then we break the quadratic down further. Let's start with this. So I'm going to do synthetic squared. long division here because it's faster. If I just do this piece, then the numbers that are going to go in here are the coefficients of my cubic polynomial, 1, 3, negative 36, and 32. And this is going to be a minus 4. Okay, so I'm going to bring the 1 down. 1 times negative 4 is negative 4. And then I subtract. 3 minus negative 4 is 7. 7 times negative 4 is negative 28. 36 minus negative 28 is negative 8. Negative 8. Okay, negative 8 times negative 4 is 32. 32 minus 32 is 0. And that kind of makes sense because, again, if we know the length, the length should be a factor of the volume. Okay, so what does that tell us? Width times height is this. x squared. plus 7x minus 8. So width times height is x squared plus 7x minus 8. What could, in terms of x, the other two side lengths be? Exactly. x plus 8 times x minus 1. So... The other two side lengths would be x plus 8 and x minus 1. Okay? We don't know which is which, but when you're working with a rectangular prism, it doesn't really matter because you can change the orientation and now the width becomes the height and the height becomes the length and all that kind of stuff, right? This, by the way, is my cell phone jail. Sometimes I use it. Anyway, it's a nice rectangular prism. Okay, so far so good. Okay, let's talk about the remainder theorem. This is just going to get easier and easier, by the way. We are just starting with the theory to try and get an understanding. of how this all works, but there are lots of shortcuts that we can take. Okay, so here's what the remainder theorem states. The remainder theorem states that if p of x is divided by x minus a, the remainder... is Q of A. Okay? This is quite nice, because now, to get the remainder, we don't even have to do long division. We can just sub the A value into the polynomial, and we get our remainder. Now, if you want the quotient, you still need to do the whole long division. division thing. But if you only want the remainder, we can do this. Okay. Now I know it seems like I just dropped this out of the sky. What I want to do though is practice doing this a few times and then I'm going to show you why it's true. So here's the idea. What is the remainder of p of x equals 2x cubed minus x squared minus 4x minus 4? What is the remainder when that is divided by x plus 3? 3? I don't know. OK. So this is our x minus a. What's the a value? x plus 3 is equal to x minus a. Yes, a is negative 3. So according to the remainder theorem, the remainder should be p of negative 3. So we're going to sub in a negative 3. And we will get, let's see, okay, so this is negative 27 times 2, negative 54, minus 9, plus 12, minus 4. Okay, that is, is that negative 55? I think it's negative 55. But I'm notorious for doing this kind of stuff wrong. Now again, this is limited because we don't know what the quotient is. We just know what the remainder is. To check that it works, let's do synthetic division quickly to make sure that that's what the remainder that we get is. So if I check with synthetic division, I'm going to do a little bit of a 2, negative 1, negative 4, negative 4. This has to be... Okay, so here is where I was talking about changing the sign or not. For the remainder theorem, the remainder theorem applies when you're dividing by a binomial of the form x minus a. So the a value is kind of going to be the opposite sign because the negative sign is built in. So I'm going to not change the sign. I'm going to make this a 3. but then I'm going to subtract. If this was a negative 3 instead, then all we would do is instead of subtracting these, we would add them. So you're kind of changing the sign twice. You end up in the same place. All right, so I'm going to bring this down, and I get 2. 2 times 3 is 6. 1 minus 6 is negative 7. Negative 7 times 3 is negative 21. Negative 4 minus negative 21. 21 is 17, right? Okay. 7 times 3 is 51. Negative 4 minus 51 is negative 55. Okay. That's the same remainder that we got. Okay. Now, this does not prove anything. All that this checking shows is that it works for this case. So we still have some work to do because I want to show you why it works all the time. But so far, okay, it works out in this scenario. Okay, all right, let's try again. Okay, so what is the remainder when this guy is divided by x minus 1? Okay, we're not plugging in a 0 though, so you're looking at this, right? At the constant. I don't know. I can't do it in my head. But according to the remainder theorem, what do we need to look at? If we were to sub in a number into this dividend for x, what would you sub in? Not negative 1. 1. Yes. So the remainder should be p of 1. So that's 1 cubed plus 2 times 1 squared minus 3 times 1 minus 1. And what is that? 3. Yeah, it is negative 1. Isn't it? Yeah. OK. Again, not helpful if we also want to know what the quotient is. But if we're just looking for the remainder, this is pretty quick. Last one. What is the remainder when 11x minus 4x to the 4 minus 7 is divided by x minus 3? What are we going to sub in? Three, p of three. Now, just so you know, if you are doing this one using either synthetic division or long division, watch out because the leading term here is this one. And there's a lot of terms missing. So we'd have to... put in a zero for the x cubed and for the x squared term. Okay, so just be aware you may have to reorder things. Okay, yeah, but this is going to be p of 3, which is 11 times 3 minus 4 times 3 to the 4 minus 7. So that's, let's see, 33. Okay, 4 times 81. 324? 4 times 80 is 320. 4 times 81 is 324. But you could do that on a calculator. Minus 7. I don't know what that is. Negative 300 and something. Or is it 200? What is it? Negative 298. Okay. All right, so now the question is, why does this work all of the time? Okay, so I do want to explore that. I'm going to do it with this question because we've already done the division. Okay, so I'm going to have to copy it onto a new page. Can you guys help me out and remember all of these things? So we're doing, it was 2x cubed, and then what was the rest of it? Minus 2x squared. It's the first one we did. Yeah, so minus 2x, no, minus x squared. x squared. Subtract 4x. Okay, and what were we dividing by? x plus 3. x plus 3. Okay, and then when we did the synthetic division, we got these coefficients. So that quotient is 2x squared minus 7x plus 17. Okay? Minus 7x plus 17. That was the quotient. Okay. And then there was the remainder of minus 55 over x plus 3. Okay. So far so good? Okay. Let's multiply the whole thing through by x plus 3. If I multiply the whole thing by x plus 3, here it cancels. So on the left side, I have 2x cubed minus x squared minus 4x minus 4. Shh. equals, this has a denominator of 1. So this is where I'm going to get that product of the x plus 3 times this, 2x squared minus 7x plus 17. And then the remainder of negative 55. Okay. So the reason why it works comes from this format. If we put in a negative 3, if we sub in negative 3 in here, what does this equal? Zero. It doesn't matter what this is. Zero times that is zero. All that's left is the remainder. Does that make sense? Yeah? You guys aren't looking so... No? You're good? Okay. I wanted to look at this through an example, but, you know, in general, if you think of, on the one hand, you have p of x equals q of x times x minus a plus the remainder. That's kind of the format that we're in. OK. If you put in p of a, which is what we were looking at, it's going to be q of a times a minus a plus r. Well, this part's 0. So all that's remaining is the remainder. Okay. Now, you don't need to know why it works, but, you know, you do and you don't because remember that... We look a lot at understanding the concepts, right? So if you can plug the numbers, that's great, okay? But ultimately, you're going to have a higher level of understanding, be able to answer more difficult questions. solve more difficult problems if you know where it comes from. I wouldn't ever ask you to derive this or come up with it yourself, but I really don't like when I'm teaching to just drop something from the sky. Here's a theorem. Use it. Let's not talk about it anymore. Okay? Questions? Okay. So I'm going to end there for today.

Um, shh. Okay, and the thing that I really wanted to sort of remind you of is firstly the process of taking a number, dividing it by a number, how we do that with long division, and then also the remainder. Okay, so, you know...

We have up to, I'm getting really distracted when people are coming in late and then catching up with, like, I get it. But if you come in five minutes late, there were people here who were on time, and that's not fair. So, you come in late, not a problem.

Just sit down and be quiet. Thank you. Okay. Okay, so...

We looked at this one example. I can't remember what the number was. It was divided by four. But, you know, once you get beyond, I don't know, like grade four or five, you stop really thinking about, you know, let's say I did 22 over three. We don't really talk about it in terms of division.

We think of it more in terms of a fraction, which is division. So what we would do with that, you know, from most of the time anyway, I don't know. from maybe, I don't know, grade five or six on, is we'd maybe convert it into an improper fraction.

We may convert that into a decimal. It'd be 7.3333333. But I just kind of want to go back to grade three, four, whatever. Okay.

And what you would have looked at before you knew what an improper versus a proper fraction is, before you knew what a repeating decimal was, you may not even know. known what a decimal was at that point. Okay?

And what we would have done is said, okay, well, if I'm doing 22 divided by 3, how many times does 3 go into 22? How many whole times does 3 go into 22? How many whole groups of 3 can you make if you have 22?

Seven. Seven. And then what's left over? One third, yeah. Right?

And essentially, this is kind of the relationship between an improper fraction and a mixed number. Okay? Now, when we're doing it with something like this, that's a much bigger number. to work with. And so, yeah, we may have to do long division to be able to get to that mixed number stage.

Okay. But this is kind of essentially where we all started. Okay.

Now, if I take this, equation and let's say I multiplied the entire thing by 3, then another way of writing the relationship between all of these numbers, so I multiply this side by 3 I have 22, if I multiply this by 3 I have 21. I'm actually going to write that as 7 times 3. Plus, if I multiply that by 3, I have 1. And this is true. If I have 22, I've got 7 groups of 3 plus an additional 1. So now what I want to do is actually introduce some vocabulary around all of this stuff. When I take 22 and I divide it by 3, the number that I have originally, this 22, is called the dividend.

Okay, the three is the divisor. Okay, the seven, what do you think that's called? It's called the quotients. And then this one.

is the remainder. Okay. Now, when we start to do polynomial division, we're going to see a similar format, either in this first line or possibly like this as well. Okay, so let's talk about that. Next.

Okay. So we've done a review of long division already. Polynomial division, just to put the last example, 22 over 3, with all of that vocabulary into context right now. What we're going to be doing here is taking some polynomial expression, and I'm going to call that p of x. And I'm going to divide it by another polynomial.

That's the divisor. I'm going to call that d of x. Okay.

Now what I end up with is some quotient, q of x, plus the remainder over the divisor. R of X over D of X. And most, almost always, that remainder is actually a constant.

It's a numerical value, okay? So just so that you can kind of see where this all came from, it's like if I put an example in here, 22 over 3 equals 7 plus 1 third. Except instead of having numbers, these are going to be...

Now, you may sort of wonder where we're going with this. Eventually, at the end of the day, what we'd like to be able to do is take a polynomial expression that's larger than degree 2, right? Because you guys are really good with quadratic expressions, ax squared plus bx plus c, and be able to factor it.

And so we need a way to sort of start to pick apart these expressions by breaking them down into smaller parts so that we can have an expression that's completely factored, right? So think like even take the number 20 and breaking it down into prime factors. You'd go 2 times 10, you could break 10 down into 2 times 5. We're gonna be doing the same thing except, you know, it's just kind of like factoring quadratics or trinomials except we're gonna be starting with a polynomial of a higher degree.

OK? And ultimately, what that will allow us to do is that if we have a function of a polynomial with higher degree, if it can be factored, we can find the roots from the factored form. OK? All right, so polynomial division, long division, works very, very similarly actually to numerical long division. I'm going to attempt to write out the steps, but know that it'll make way more sense when you actually see an example.

So I actually don't think I'm going to write out every single step. I'm just going to write the beginning part. And the beginning part is essentially we... You set it up the same way as long division, with that sort of... half a box thing.

And what we're going to do is divide the leading term of the dividend by the leading term of the divisor. Okay, so I don't know, let's say we're working with something that's a cubic polynomial, and this is like, the leading term is 3x cubed, and then we divide it by x plus 2. We're just going to take the leading term, the 3x, 3x cubed divided by x because that's the leading term of the divisor. Okay. Once you do that, you multiply through, subtract, bring down the next one, repeat. Okay.

So that's the part where it's really hard to put it into step by step without an example beside it. So we're going to now multiply through. Subtract, bring down the next term, and then add heaps. Okay, so I know that that doesn't make a lot of sense right now, but hopefully when you see an example you'd be like oh yeah okay I kind of see what's going on here. Okay now one thing that is very important is that when you start with the dividend within the long division you have to see every power of x.

So for example if you have a cubic polynomial you have to have an x cubed term and x squared term, an x term, and a constant. And if you are missing any of those in the original, you're going to put in a placeholder. Like let's say there's no x squared term.

You're going to add in zero x squared. So it's a placeholder. Okay.

It's kind of similar to place value with numbers, right? If we have the number 3,021, we write that as 3,021 to show that there are 0 hundreds in that number. It's a similar sort of idea.

Okay? Okay, let's give this a try. Okay, so here is our polynomial, our dividend 2x cubed plus 3x squared minus 4. and we are going to divide that by the divisor x minus 1. And ultimately, the goal is to write it in this form. Okay?

So here's how we're going to start. We're going to set up a nice long division symbol. This is what I'm dividing by, so this is going to go over here, x minus 1. And then I have to put this polynomial under the division sign. Now this is cubic.

I have an x cubed term. I have an x squared term. I don't have an x term. So I'm going to add in 0x.

as a placeholder. And then I have the minus 4. So this is going to be 2x cubed plus 3x squared plus 0x minus 4. Yes? That's what we're talking about, yeah. Okay?

Alright, so first step, we take the leading term of the dividend. That would be the 2x cubed and we're gonna divide it by the leading term of the divisor, which is x. Okay, so what is 2x cubed divided by x?

2x squared. So we're gonna put that up here. Okay, now we multiply through. This times that goes there. So 2x squared times x is 2x cubed.

2x squared times negative 1 is minus 2x squared. Yes? How did you get 2x squared? I went 2x cubed divided by x.

Yeah, leading term of this divided by leading term of that. Okay? All right, so I multiply through. I'm going to subtract.

This by design is 0. That's what I want. But 3x squared minus negative 2x squared is 5x squared. Okay, now bring this down.

Plus 0x. Okay, so now I'm going to repeat. And I'm going to go, okay, 5x squared divided by x. What's 5x squared divided by x?

5x. So I'm going to add 5x. I'm going to multiply through.

5x times x is 5x squared. 5x times negative 1 is minus 5x. And I'm going to subtract.

Okay. Now again, by design, this is 0. 0x minus negative 5x is 5x. Yep. This one?

So I put a 5x over there, and now I'm multiplying through 5x times x minus 1. That goes there. Yeah. Okay.

All right. So I subtracted. I'm left with 5x.

Bring this down. Minus 4. And I'm going to do the same thing. Okay. So now I'm dividing the leading term here as 5x, dividing that by x. 5x.

Divide by x, 5. And I'm going to multiply through. 5 times x minus 1 is 5x minus 5. And when I subtract, this is 0. This is negative 4 minus negative 5 is 1. That's my remainder. Okay, so this over here is the dividend. This, divisor.

This over here, quotient. That's the remainder. Okay, now in this format, okay, so p of x was my original polynomial. Now at this stage, you can take out the 0x. If you want to put it in, that's fine, but you don't have to have it in there.

Okay, so that was 2x cubed plus 3x squared. minus 4, oh I didn't give myself very much room, I'll have to do it on the next line, over the divisor is x minus 1, equals the quotient 2x squared plus 5x plus 5, plus the remainder 1 over x minus 1. Okay, now I know that this can seem abstract, but if you remind yourself how it works with numbers, if we were to do the same thing with like 22 divided by 3, right, we would do this. Or 22, sorry. This goes in 7 times. 7 times 3 is 21. Remainder 1. So 22 over 3 equals 7 plus 1 over 3. Okay.

Yes? Is there any, like, practice sheet? Is there any white?

There's a lot of practice, yes. That's going to be your homework today. Yes, okay. So, that's polynomial long division.

Okay? Now, in reality, we have some shortcuts that we can take. And essentially, like to understand why the shortcut works, you actually have to go back to what I talked about Last class in that in fact place value is a condensed way of writing a number based on powers of 10 and So when we actually write the number out. We don't write this whole thing.

We write 3,241. 3, 4, or 3, 2, 4, 1. Okay. Now, when we did the polynomial long division, we didn't do that.

We kept all those power of x in there. Synthetic division essentially removes all of the powers of x and just keeps the coefficients. Just like when we're working with the number 3,241, we don't write it as 3 times 10 cubed plus 2 times 10 squared, et cetera. All right, now, synthetic division.

Same process, except we are taking a few shortcuts so that we don't have to write all of the powers of x down. Okay, so here's how it works. What you're going to do is set up an upside down division sign. And over here in the top line, there should be enough room for sort of two lines.

In the top line, you are going to put just the coefficients of the dividend in descending order. So I'm going to put in 2, 3, 0, negative 4. Okay, what I'm going to put here is if this is written in the form x minus a, I'm going to write just the negative one piece. Okay.

So now, here's what we're going to do. Bring this down. Multiply.

2 times negative 1, negative 2. Okay? Now, subtract. 3 minus negative 2 is 5. Multiply. 5 times negative 1 is negative 5. 0 minus negative 5 is 5. Multiply again. 5 times negative 1 is negative 5. Subtract.

4 minus negative 5 is 1. Okay, so now we're done. Now here's how we interpret what we ended up with. This is the remainder.

These three numbers, 2, 5, 5, make up the coefficients of the quotient. Okay, so it was 2x squared plus 5x plus 5. Remainder 1. Okay, so again... 2x cubed plus 2x, sorry, 3x squared minus 4 is equal to, okay, that's the quotient.

2x squared, sorry, over x minus 1. 2x squared plus 5x plus 5. And then the 1 is the remainder. Plus 1 over x minus 1. Okay. Now, I do want to warn you, if any of you guys have tutors or look at YouTube videos or whatever, there are two ways to do synthetic division. Well, it's all pretty much the same, but here's how things differ. Some people keep this sign.

and maintain it. If you do that, you subtract. Some people change that to a positive one, and then instead of subtracting, they add.

Okay? You end up with the same quotient and remainder at the end of the day. Okay?

I was actually taught to change the sign and add, but I kind of like the idea of not changing the sign and subtracting because it It models the long division process more closely. But you guys can do whatever you want. Yes? If it was x plus 1, would you add?

No, you always subtract. The question is, do you change this side? sign from the x minus a? Do you keep that negative one or do you change it to a one?

Okay. Now I'm going to stick with this where I don't change the sign. I subtract because again, in long division, that's what you do. You subtract. But I just want you guys to know that I would say it's a pretty much 50-50 split between who does it change in the sign and adding and who keeps the sign and subtracting because otherwise you guys are going to come to me and be like, wait, how come these people are subtracting?

are adding instead of subtracting, okay? And you can do it either way. Yeah?

I know we just did that question. Yes. Yes.

Yes. How do you know it's 2x squared? Okay, good question.

So what you want to keep in mind is that in the first step of long division, you're dividing this leading coefficient by this guy. So this is always going to be one. degree less than whatever that is.

Okay, all right, so let's try this. Now you guys are sort of expected to know both processes, writing out the full polynomial long division and also just writing out synthetic division. Ultimately though most people just do synthetic division because it's a lot faster. Yes? Yes.

Okay. Now again, where this is leading, like we're not just doing this for fun. Ultimately, where this is going is here.

Let's say we get a remainder of zero. What does that mean? If I did 21 divided by 3 instead of 22 divided by 3, what's my remainder? 21 divided by 3? Zero.

Okay, so what does that tell me about the 3 and the 21? Yeah, it's a factor, exactly. So if we end up with a remainder of 0, what we know is that, OK, we have a factor.

And so that's going to help us in the factoring process. Okay, now we still have lots to learn because all, you know, if you're just trying to see is something a factor, I mean, you kind of want to have a starting place and know, okay, well, what should I be testing out here? We'll get to that.

But right now we're just learning this process. Okay. All right.

Let's try this guy. Okay, so we have x cubed plus 7x squared minus 3x plus 4 divided by x minus 2. So I am going to set up this upside down division sign. And over here, I'm going to put that negative 2 value. Okay.

So within here on the top line, I'm going to put the coefficients of my evident. And I'm going to make sure that I'm not missing any powers of x. Because if I'm missing a powers of x, I have to put a zero in as a placeholder.

But it looks like I've got everything there. So I've got 1, 7, negative 3, and 4. And I'm going to start by bringing this down. Okay?

And what we do now is multiply, subtract. Multiply, subtract. Multiply, subtract.

Yes? A double, like if this was 14. Oh, if the coefficient was double digits, you just write them both down. Yeah. So in that way, I guess like it's not the same as place value because with place value. if you had 14 groups of 10 let's say you wouldn't write it that way you would one group of 100 and four groups of 10 that doesn't carry through there are some holes in the modeling so 1 times negative 2 is negative 2 subtract what do we get?

9 9 9 times negative 2, negative 18. Okay, subtract. 15. 15 times negative 2, negative 30. Subtract. Yes.

For some reason in my head I had 26. 34. Okay. So in terms of, I mean, this is just a, like a graphic organizer to get us to our answer. This is not the answer though.

Okay. Ultimately, the answer is... When we take this original polynomial, minus 3x plus 4, and divide it by x minus 2, we get a quotient of, this is 1x squared plus 9x plus 15. And then here's our remainder, plus 34 over x minus 2. Can you just do synthetic division?

Okay, so the short answer is for this course, you need to be able to show both. In the long run, does anyone really do the full long division? No, because this is an efficiency, right?

Yeah. Okay. So one other thing that I just want to show you is that this is one format that you could be asked for.

Another way that you might see this being written is where you essentially Multiply the entire equation by the x minus 2 so that there's no denominators. And that was kind of like when I did, did I do it here? No, but I did 22 equals 7 times 3 plus 1. So how did I get that? I multiplied the whole thing by 3. So you could see that too. If I multiply the whole thing by x minus 2, this denominator is going to be gone.

False. I'll have x cubed plus 7x squared minus 3x plus 4. This is the only one that doesn't have a denominator of x minus 2. So that's where I'm going to see the x minus 2 times this quotient. And then here I have this remainder of 34. What's that? Oh, okay, so what we were solving was a rational equation. It's a similar idea in that you're multiplying through by the least common denominator to isolate x.

except that we're not solving for a value of x, really, that satisfies the equation. We're more manipulating the algebra, so it looks a different way. Yeah, but it's a similar idea in terms of multiplying through by the least common denominator.

Okay? All right. Let's try this one. Okay, so where do I need to start?

Yeah, sure. Let's do the long way. We can do both. Okay, so I have to rewrite it, though, for the long way. It's 3x squared.

No, 3x cubed. And what is it? Plus 8x? Minus 8x. Minus 1?

Okay. Over x plus 2. Okay? All right, so the long way... is I have to write it like division, right?

So this is x plus 2. And then, but again, just like synthetic division, I still need a placeholder for the x squared term. So I'll write this as 3x cubed plus 0x squared minus 8x minus 1. Okay, and then, so what do I do first? Yeah, so take leading term divided by leading term. Okay, so 3x squared divided by x is 3x cubed, thank you, is 3x squared. Then I'm going to multiply through.

So 3x squared times x plus 2 is 3x cubed plus 6x squared. And then I subtract. Okay, so this piece is 0. This is negative 6x squared. Then I bring this down, minus 8x. Okay, so what do I do next?

Yeah? Almost. What do I need to multiply through, though, by? Yeah, I need to divide.

Yeah, so I'm going to go 6x, negative 6x squared divided by x. And that's minus 6x. Okay, and now I can multiply through. So negative 6x times x is negative 6x squared.

And negative 6x times 2 is minus 12x. Okay, and now I can subtract. Watch out for the double negatives, though, when you subtract, right? That's one place where sometimes people make careless mistake. Okay, so...

Negative 8x minus negative 12x is 4x minus 1. And then the last piece, 4x divided by x is 4. And then 4 times x plus 2 is 4x plus 8. And we end up with... Negative 9. So that's our remainder. Now this is also different than how it would work with numbers because when we're working with numbers, our remainders will only be positive. Whereas in this case, what we're actually looking for is a constant. Okay?

So. At the end of the day, how would we show the relationship between, there's four things, right? There's the dividend, the quotient, sorry, the divisor, quotient, remainder. And if you ever forget... where everything goes, do the 22 over 3 equals 7 plus 1 third.

And that will help remind you. Okay? All right.

So we're going to have 3x cubed minus 8x minus 1. over x plus 2 equals the quotient, which is 3x squared minus 6x plus 4 plus the remainder. So minus 9 over x plus 2. Okay, now just so that you can see the synthetic division and sort of follow it through this problem, if we were to do this with synthetic division instead, then I would still have those same coefficients, okay? This would be a 2. I'm bringing this down.

Multiplying, 3 times 2 is 6. Subtracting, negative 6. Okay. Multiplying, negative 6 times 2 is negative 12. Subtracting, 4. 4 times 2 is 8. Subtracting is, oh, sorry, this is a minus 1. Negative 9. Okay. So. We kind of see like, okay, and then we have this and that. I'm trying to like color code it, but this and that, this and that.

Okay, and then we have. The negative 12 and 4, and then the 4 and the 1, the numbers are all the same. We're just kind of...

We're dropping the powers of x so that we don't have to keep track of those as well. Okay. All right. So.

Here's an example. The volume of a rectangular prism is given by v of x equals x cubed plus 3x squared minus 36x plus 32. Determine possible measures for the width and the height in terms of x if the length is x minus 4. Okay. So...

Let's just remind ourselves this is a rectangular prism. How do we get the volume of a rectangular prism? Yeah. Oh, I didn't do that. Volume is length times width times height.

So what we are after is what could these two be? What could the width be and what could the height be? If we know the volume and the length, let's start out by just, I mean, what would W times H be? Let's just start with that. This part.

What if I isolated the width times the height? What would you have to do to get- this piece by itself. Right now it's being multiplied by length, divided by length. So if I divide both sides by length, then whatever I get is the width times the height.

So an equivalent type of question with numbers would be, okay, you have a rectangular prism, the volume... is 30. And the side lengths are all whole numbers. Okay?

So now if I tell you the length is 5, what could the other two sides be? The volume is 30, the length is 5. 3 and 2, right? Because the width times the height would have to be 6. 5 times width times height is 30. 5 times 6 is 30. So the width times the height would have to be 6. to be 6. So it could be 3 and 2. It could also be 1 and 6. Similar type of question.

All right. Well, let's start with just trying to get an expression for width times height. Width times height is going to be the volume x cubed plus 3x squared minus 36x plus 32 over x minus 4. OK, now, x minus 4 should be a factor, because we're told that it's the length. And the length should be a factor of the volume.

So ultimately, if we divide these, we're dividing a cubic polynomial by a linear. polynomial, we should end up with something quadratic. If we have something quadratic, the hope is that we can factor it to get the other two sidelines.

So we're starting with cubic, breaking it down into a binomial times a quadratic, then we break the quadratic down further. Let's start with this. So I'm going to do synthetic squared. long division here because it's faster. If I just do this piece, then the numbers that are going to go in here are the coefficients of my cubic polynomial, 1, 3, negative 36, and 32. And this is going to be a minus 4. Okay, so I'm going to bring the 1 down.

1 times negative 4 is negative 4. And then I subtract. 3 minus negative 4 is 7. 7 times negative 4 is negative 28. 36 minus negative 28 is negative 8. Negative 8. Okay, negative 8 times negative 4 is 32. 32 minus 32 is 0. And that kind of makes sense because, again, if we know the length, the length should be a factor of the volume. Okay, so what does that tell us?

Width times height is this. x squared. plus 7x minus 8. So width times height is x squared plus 7x minus 8. What could, in terms of x, the other two side lengths be?

Exactly. x plus 8 times x minus 1. So... The other two side lengths would be x plus 8 and x minus 1. Okay? We don't know which is which, but when you're working with a rectangular prism, it doesn't really matter because you can change the orientation and now the width becomes the height and the height becomes the length and all that kind of stuff, right?

This, by the way, is my cell phone jail. Sometimes I use it. Anyway, it's a nice rectangular prism.

Okay, so far so good. Okay, let's talk about the remainder theorem. This is just going to get easier and easier, by the way.

We are just starting with the theory to try and get an understanding. of how this all works, but there are lots of shortcuts that we can take. Okay, so here's what the remainder theorem states.

The remainder theorem states that if p of x is divided by x minus a, the remainder... is Q of A. Okay? This is quite nice, because now, to get the remainder, we don't even have to do long division.

We can just sub the A value into the polynomial, and we get our remainder. Now, if you want the quotient, you still need to do the whole long division. division thing. But if you only want the remainder, we can do this.

Okay. Now I know it seems like I just dropped this out of the sky. What I want to do though is practice doing this a few times and then I'm going to show you why it's true.

So here's the idea. What is the remainder of p of x equals 2x cubed minus x squared minus 4x minus 4? What is the remainder when that is divided by x plus 3? 3? I don't know.

OK. So this is our x minus a. What's the a value?

x plus 3 is equal to x minus a. Yes, a is negative 3. So according to the remainder theorem, the remainder should be p of negative 3. So we're going to sub in a negative 3. And we will get, let's see, okay, so this is negative 27 times 2, negative 54, minus 9, plus 12, minus 4. Okay, that is, is that negative 55? I think it's negative 55. But I'm notorious for doing this kind of stuff wrong. Now again, this is limited because we don't know what the quotient is. We just know what the remainder is.

To check that it works, let's do synthetic division quickly to make sure that that's what the remainder that we get is. So if I check with synthetic division, I'm going to do a little bit of a 2, negative 1, negative 4, negative 4. This has to be... Okay, so here is where I was talking about changing the sign or not. For the remainder theorem, the remainder theorem applies when you're dividing by a binomial of the form x minus a.

So the a value is kind of going to be the opposite sign because the negative sign is built in. So I'm going to not change the sign. I'm going to make this a 3. but then I'm going to subtract. If this was a negative 3 instead, then all we would do is instead of subtracting these, we would add them. So you're kind of changing the sign twice.

You end up in the same place. All right, so I'm going to bring this down, and I get 2. 2 times 3 is 6. 1 minus 6 is negative 7. Negative 7 times 3 is negative 21. Negative 4 minus negative 21. 21 is 17, right? Okay. 7 times 3 is 51. Negative 4 minus 51 is negative 55. Okay.

That's the same remainder that we got. Okay. Now, this does not prove anything.

All that this checking shows is that it works for this case. So we still have some work to do because I want to show you why it works all the time. But so far, okay, it works out in this scenario.

Okay, all right, let's try again. Okay, so what is the remainder when this guy is divided by x minus 1? Okay, we're not plugging in a 0 though, so you're looking at this, right? At the constant. I don't know.

I can't do it in my head. But according to the remainder theorem, what do we need to look at? If we were to sub in a number into this dividend for x, what would you sub in? Not negative 1. 1. Yes.

So the remainder should be p of 1. So that's 1 cubed plus 2 times 1 squared minus 3 times 1 minus 1. And what is that? 3. Yeah, it is negative 1. Isn't it? Yeah. OK. Again, not helpful if we also want to know what the quotient is.

But if we're just looking for the remainder, this is pretty quick. Last one. What is the remainder when 11x minus 4x to the 4 minus 7 is divided by x minus 3? What are we going to sub in?

Three, p of three. Now, just so you know, if you are doing this one using either synthetic division or long division, watch out because the leading term here is this one. And there's a lot of terms missing.

So we'd have to... put in a zero for the x cubed and for the x squared term. Okay, so just be aware you may have to reorder things. Okay, yeah, but this is going to be p of 3, which is 11 times 3 minus 4 times 3 to the 4 minus 7. So that's, let's see, 33. Okay, 4 times 81. 324?

4 times 80 is 320. 4 times 81 is 324. But you could do that on a calculator. Minus 7. I don't know what that is. Negative 300 and something.

Or is it 200? What is it? Negative 298. Okay. All right, so now the question is, why does this work all of the time?

Okay, so I do want to explore that. I'm going to do it with this question because we've already done the division. Okay, so I'm going to have to copy it onto a new page.

Can you guys help me out and remember all of these things? So we're doing, it was 2x cubed, and then what was the rest of it? Minus 2x squared.

It's the first one we did. Yeah, so minus 2x, no, minus x squared. x squared. Subtract 4x.

Okay, and what were we dividing by? x plus 3. x plus 3. Okay, and then when we did the synthetic division, we got these coefficients. So that quotient is 2x squared minus 7x plus 17. Okay? Minus 7x plus 17. That was the quotient. Okay.

And then there was the remainder of minus 55 over x plus 3. Okay. So far so good? Okay.

Let's multiply the whole thing through by x plus 3. If I multiply the whole thing by x plus 3, here it cancels. So on the left side, I have 2x cubed minus x squared minus 4x minus 4. Shh. equals, this has a denominator of 1. So this is where I'm going to get that product of the x plus 3 times this, 2x squared minus 7x plus 17. And then the remainder of negative 55. Okay. So the reason why it works comes from this format.

If we put in a negative 3, if we sub in negative 3 in here, what does this equal? Zero. It doesn't matter what this is.

Zero times that is zero. All that's left is the remainder. Does that make sense? Yeah? You guys aren't looking so...

No? You're good? Okay. I wanted to look at this through an example, but, you know, in general, if you think of, on the one hand, you have p of x equals q of x times x minus a plus the remainder. That's kind of the format that we're in.

OK. If you put in p of a, which is what we were looking at, it's going to be q of a times a minus a plus r. Well, this part's 0. So all that's remaining is the remainder. Okay.

Now, you don't need to know why it works, but, you know, you do and you don't because remember that... We look a lot at understanding the concepts, right? So if you can plug the numbers, that's great, okay? But ultimately, you're going to have a higher level of understanding, be able to answer more difficult questions.

solve more difficult problems if you know where it comes from. I wouldn't ever ask you to derive this or come up with it yourself, but I really don't like when I'm teaching to just drop something from the sky. Here's a theorem.

Use it. Let's not talk about it anymore. Okay? Questions?

Okay. So I'm going to end there for today.

Transcript for:Understanding Long Division and Polynomial Division

Transcript for:
Understanding Long Division and Polynomial Division