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Welcome to Algebra I: Dividing Polynomials

31 minutes

Hey, guys, welcome to Algebra I. Today's lesson focuses on dividing polynomials. All you know about factoring and the laws of exponents will get you through this lesson. You ready? Let's go. All right. Let me jog your memory first. Do you remember how to simplify something like this? If we apply the laws of exponents to simplify 20x to the 5th over 4x squared-- Let me be sure I've got my pen here. To handle this one, the first thing that we do is handle the coefficients. We'd go ahead and divide 20 divided by 4, and say, "All right, that's 5." Then x to the 5th divided by x squared-- the laws of exponents say, in a situation like this, a quotient of powers, subtract the exponents. So 5 minus 2, that's 3. The other part is x to the 3rd. We did that earlier in the course. That helps us learn how to divide polynomials. Look at this one. Deep breath 'cause that's a long one. We've got 35y to the 4th plus 20y squared minus 18y-- that whole quantity-- we're going to divide it by 5y. It looks a little intense, but it breaks down to something simple. We're going to take each of these terms and divide it by 5y. I'll rewrite this differently. It might make you more comfortable if you saw it like this: 35y to the 4th divided by 5y plus 20y squared divided by 5y minus 18y divided by 5y. That's all that we'll do here. Essentially, you are taking that 5y and dividing each term by it. Whatever our answer is is the answer. Okay? I need more room to work. Let's get rid of some stuff. Let's get rid of example one. Let's cut that out of the picture. Let's move this up, and then let's move this up so we can get more space up here. Now let's get our pen back, and we're really to rock. If we look at this first chunk right here, we'll handle coefficients first, like you already know. You ignore the rest of this problem and just focus your attention on that first bit. So 35 divided by 5, I know that's 7. Now, y to the 4th divided by y. Remember, there's an invisible 1 we don't see, right? To handle y to the 4th divided by y to the 1st, subtract exponents. That'd be y to the 3rd. That first part just simplifies to 7y to the 3rd. Let's look at the second part. It's plus, so put that in there-- 20y squared divided by 5y to the 1st. Get that in there. What are we're doing first? Right, handle those coefficients-- 20 divided by 5, that's 4. y squared divided by y to the 1st, subtract the exponents. 2 minus 1, that's 1. This is y to the 1st, but when it's a 1, we don't typically write the exponent, so 4y for the second one. Let's look at the last one. We have 18 divided by 5. That might have made you uncomfortable because it's not dividing evenly, but we're not afraid of fractions in Algebra I. I can't simplify this to anything simpler than 18 divided by 5, so I'll leave it as 18 over 5. Just leave it like a fraction. I've got y to the 1st, right, divided by y to the 1st-- so 1 minus 1, that's 0. Think back-- I'm coming off to the side. Remember, anything to the 0 power is 1. So I'm going to be multiplying 18 divided by 5 times 1, but that doesn't change anything. Leave it as is. Just leave 18 divided by 5, and you've got it. This polynomial is your answer. When you divide what we started with divided by 5y, that's what you get.

(female describer) 7y cubed plus 4y minus 18 over 5.

(teacher) That's your answer for that one. Let's try another one. Find the quotient. We have 18x to the 6th plus 27x to the 5th minus 45x cubed, all divided by 9x squared. This problem is set up differently, but you are doing the same thing. You're going to find the quotient. I break this into chunks, like we just did. In math, if you can chunk things up and look at them in bits, it's easier to process a problem that way. You can manipulate this the way you want to get through it. As long as you follow the rules for math, do whatever you want. I get that out of our way. Better move that up a bit; get some room. I'm going to chunk this problem up so it's easier to process. I'm going to write it as 18x to the 6th divided by 9x squared. Then I have plus... 27x to the 5th divided by 9x squared. Then I have minus 45x to the 3rd divided by 9x squared. I've got that all broken up. I'm going to handle each piece separately and get through it. If I look at the first bit, the coefficients first, so you have 18 divided by 9-- that's 2. I bet your mental math is good now that you've been working with these numbers. Now look at that variable part-- x to the 6th divided by x squared. So what do we do there? Subtract the exponents. So, 6 minus 2, that's 4. So, 2x to the 4th. I'm done with that first part. I'm crossing it out. Look at the middle bit. Look at our coefficients. I'm bringing the plus sign down-- 27 divided by 9, that's 3. And then as far as our exponents go, what will we do? Subtract. So, 5 minus 2, that's 3. This part's going to be x cubed. Let's look at the last part. I brought my minus down. So, 45 divided by 9, that's 5. Okay? And I handle my exponents over here. I need to subtract-- 3 minus 2, that's 1. When it's an exponent of 1, we typically don't write it. That crazy-looking problem that started out simplifies to this-- 2x to the 4th plus 3x cubed minus 5x. And we're all done. Look at the next one.

(female describer) Example 3: Find the quotient. 20a to the 6th, b to the 4th minus 16a to the 4th, b squared, all over 4a cubed, b. All right, so this one's different because we've got two different variables in this problem, but the process is still the same, okay? I told you that you'll never run into the same type of problem every time, so don't memorize exactly the steps I'm taking, but memorize the process. It's the same process I'm taking every time. What I do here to have room to work is get rid of stuff. I'm going to slide that up. The first thing I do every time is chunking it out, just breaking it up into chunks. This one's short, so I'm not going to have but two chunks. I'll have 20a to the 6th, b to the 4th divided by 4a cubed b. Got the first bit. Then I have minus 16a to the 4th, b squared over 4a cubed b, okay? Got it into chunks. I simplify what I've got here. Same thing--handle your coefficients first. You have 20 divided by 4-- mental math--that's 5. If you don't want to do mental math, pull out a calculator. Handling the "a"s, I have 6 and 3, so I need to subtract those. 6 minus 3 is 3, so a to the 3rd. Then I have a b, so I know it's an invisible 1. Subtract these. So 4 minus 1, that's also 3, so b cubed. That first part just simplified down to 5a cubed, b cubed, and you've handled that. You're done with that. Bring down the minus sign, then we handle this part. Coefficients first-- 16 divided by 4, that is 4. Let's handle these exponents with the "a"s. We've got 4 and 3, I need to subtract. 4 minus 3, that's 1. So, a to the 1st, but we don't generally write the 1. If I handle my "b"s-- there's an invisible 1 to help me subtract. 2 minus 1, that is 1. So, then I've got just b. There you go for that one. That quotient just simplified to 5a cubed, b cubed minus 4ab. Let's keep going. What do you know? It is your turn. Press pause, take a few minutes, and work through these problems, and when you're ready to compare answers, press play.

(female describer) Find each quotient. 1. Open parenthesis, 28x to the 7th minus 49x to the 6th plus 21x to the 4th, close parenthesis, divided by 7x cubed. 2. 30x to the 5th, y cubed plus 50x to the 4th, y squared, all over 10x squared, y squared.

(teacher) Let's see how you did. Let me get my pointer to move these out of the way. For the first one, your answer should have been 4x to the 4th minus 7x cubed plus 3x. And for the second one, 3x cubed y plus 5x squared. Now, if you want to see how I got those, I'll show you. Okay. I followed the same process we've been doing. I started out by rewriting this into my chunks, into those three chunks. At first--I'm going to need a little more room. Get this out of the way, and let's move that up. Here we go. Get my chunks going. 28x to the 7th divided by 7x cubed minus 49x cubed-- I'm sorry, x to the 6th-- divided by 7x cubed plus 21x to the 4th divided by 7x cubed. Okay? I've got it in my chunks, now I simplify. I handle my coefficients first-- 28 divided by 7, that's 4. Subtract my exponents here. So 7 minus 4, that's 4-- I'm sorry, 7 minus 3-- that's 4, all right, minus 49 divided by 7. That's 7. Subtract my exponents, and I get 3. All right, got that chunk. Last bit, got my plus sign. 21 divided by 7 is 3. Subtract my exponents, and I get 1. We don't typically write the exponent when it's 1. There we go. That is how I got the first one. If you want to see the second one, here we go. Same process. Break it up into chunks. Let's move some things out of the way. All right, let me get my chunks. I have 30x to the 5th, y cubed, and I divide that by 10x squared, y squared-- I've got the first bit-- plus 50x to the 4th, y squared, divided by 10x squared, y squared. Let's just break it down: 30 divided by 10 is 3. Subtract my exponents with my "x"s. I'd get x to the 3rd. Subtract my exponents with my "y"s, and I'd get y to the 1st, which, remember, we don't generally write. I've handled that. Plus 50 divided by 10-- that's 5. Subtract the exponents, so 4 minus 2, that's 2. So, x squared-- 2 minus 2, that's 0. Remember, anything to the 0 power is 1, so essentially, it would be 5x squared times 1, which doesn't change anything, so just disregard it. That's how we got the second one-- 3x cubed, y plus 5x squared. All right. There's a little more to this. There's also a bit of this where you have to know how to factor to find how to divide some polynomials. To jog your memory, let me bring you back to fractions. When we simplify fractions, basically what we're doing is finding the greatest common factor and canceling it out. For example, 12 divided by 15-- I find factors of 12 and 15, and I want there to be a greatest common factor between them. I'm thinking about 12 and 15 and about the factors. The pair that's going to work for me-- I know 4 times 3 is 12, and I know that 5 times 3 is 15, so 3 is the greatest common factor here. You can cancel out 3s, which would leave you with 4 over 5. You could say 12/15 simplifies to 4/5 because you factored the 12, you factored the 15, and then you canceled out the greatest common factor. Keep that process in mind because we're going to do the exact same thing but with polynomials.

(female describer) Example 4: Find the quotient. x squared minus 2x minus 3, all over x minus 3. We're going to factor that numerator. Our denominator already has one factor, and we're going to cancel some things out to get through this one. Think about factoring because we need to factor x squared minus 2x minus 3. I come down--let's go here and do scratch work. I'm trying to factor this. So, if you remember, you're looking for a pair of factors-- First, factor the 3. Then you want to find the pair that the sum is negative 2. Let's factor 3. I'm ignoring the sign to start with. 1 times 3--3 is prime, so that's a short list-- 1 times 3 gives me 3. Now consider your signs. You really need a negative 3 and you need to combine these to get a negative 2. When I'm multiplying, if my answer is negative, I've multiplied a negative number times a positive number. I figure which one is going to be negative so that when I combine them, I also get negative 2. What if the 1 was negative? If 1 was negative, negative 1 times 3 gives me negative 3, but negative 1 plus 3 gives me positive 2, and I need negative 2. That's telling me, okay, the 1 is not what's negative here. It'll have to be 3 that's negative, because 1 times negative 3 will give you negative 3, and 1 plus negative 3 is negative 2. This trinomial will factor to x plus 1, x minus 3. Now I'm going to go back to my problem, rewriting that numerator by its factors instead. I replace x squared minus 2x minus 3 with the factors, x plus 1, x minus 3. Let's do a little moving around here, get some space. I replace that numerator with x plus 1, x minus 3. I'll get that scratch work out of our way so things don't get too cluttered up here. I've replaced the numerator and written it by its factors. My denominator is still my denominator-- still x minus 3. Now I step back and look at it. Notice a factor in your numerator that's also in your denominator? The x minus 3. Like the fractions, where we crossed out those factors, we're going to cross out those common factors. The answer is just this factor I'm left with, this x plus 1. And you're all done. That's what it was here. With these problems, the part that takes a little work is just factoring the numerator. After you've factored your numerator, it's smooth sailing. Cross out the common factors. Let's try the next one. Our numerator-- we're finding a quotient. We've got x squared plus 7x plus 10. We divide it by x plus 5. Remember, start out by factoring the numerator. I'm going to scoot that a bit. I come to the side; do a little scratch work. So x squared plus 7x plus 10. From what I know, I need to factor 10. So, for 10-- I know 1 times 10 is 10, 2 times 5 gives me 10. I look at the trinomial and consider the signs. I need a positive 10 for my product. That means I'm multiplying two positive numbers together or two negative numbers together. Looking at my trinomial, I need a sum of positive 7. I could not add negative numbers together and get something positive, so that means my factors here are going to be positive. Let's test it out. I need to find out which pairs, when added, is going to give me positive 7. I know 1 plus 10 is 11, so it's not that pair. 2 plus 5 is definitely 7, so this is the pair that works for my factors. I go to the problem and rewrite that x squared plus 7x plus 10 by its factors, x plus 2 and x plus 5. I write it here so you understand before I move over there. I'm going to replace that trinomial with its factors. That'll help me simplify this quotient. All righty. I do need to erase; I'll run into that up here. I'll leave that trinomial. I leave those factors and copy those. I write x plus 2, x plus 5, and then for my denominator, I still just have that x plus 5. I have handled the numerator; I've got the denominator. I get this out of our way. Remember what we did in the last problem-- this is where you see what you cancel out, what's common to the numerator and the denominator. In this case, it's x plus 5, so cancel those out, and you're left with x plus 2, and that is your answer. All right? Let's keep going. Let's try another. Find the quotient of x squared minus 7x plus 12. What do I do first? Right, factor the 12. I'll come off to the side, okay? I know 1 times 12 is 12. So is 2 times 6, and so is 3 times 4. And that's it. If I consider my signs here-- Let me rewrite the trinomial so we look at that by itself. Okay, so if I consider my signs-- I need to multiply and get positive 12. That means I'm multiplying two positive numbers together or two negative numbers together, but because I need a sum of negative 7, that's telling me it's the two negatives-- that's going to be the case. Run down your list and see which pair works. If these were negative 1 and negative 12, if I combine those, I get negative 13. That's not the pair. If I had negative 2 and negative 6-- if I combined those, I'd get negative 8. So that's not the pair. If I had negative 3 and negative 4, I'd get negative 7, so that's the right pair. This trinomial is going to factor to x minus 3, x minus 4. Going back to our problem, 'cause we finished our side work, we replace that numerator and write it by its factors. We write the x minus 3, the x minus 4 for the numerator. Let's scoot a few things around here, get some space. I scooted that off too far, I think--there we go. So, x minus 3, x minus 4. All right? And that's typically when things get crowded, so let me get rid of some things. All right. Then I throw my denominator in there, that x minus 4. All right, what do we do now? We get rid of stuff. We cancel that out, and you're just left with x minus 3. And you're all done. All right, I think it's about to be your turn. Take a few minutes; press pause. Work through these problems. When you're ready, press play.

(female describer) Find each quotient. 1. x squared plus 8x minus 9 all over x minus 1. 2. x squared plus 5x minus 24 all over x plus 8.

(teacher) All right, you ready? Let's see how you did. Let's get the pointer here to move these. For number one, the answer was x plus 9. That was your quotient. For number two, x minus 3. If you want to see my work, this is what I did: the process we've been doing. Start out by factoring the numerator. So I need factors of 9, and I believe I do not have my pen. So I need factors of 9, so 1 times 9 gives me 9 and so does 3 times 3. Now I consider my signs. I need negative 9, so that means in each pair, one number's negative, but I need to be able to combine those pairs and get a positive 8. Is there any way to combine 1 and 9 and get positive 8? Mm-hmm. If the 1 were negative. Negative 1 plus 9 is positive 8. That means that's the right pair of factors. Now I'm over here, back in my problem. I replace that trinomial and write it by its factors: x minus 1, x plus 9. I'll get that scratch work out of the way so it doesn't confuse you. You do not need to erase your scratch work. I do it to get a clear view of what's going on. We've got the numerator factored. Let me write the denominator in here, x minus 1. You're at the fun part-- cancel what's common: x minus 1, x minus 1. And that just left you with x plus 9. That's how I got that first one. I'll show you how I got the second one. Did the same thing. Factor 24, because I'll ignore the signs to start with. 24's got a longer list. I know 1 times 24, 2 times 12, 3 times 8, and 4 times 6. That completes the list for the factors of 24. We need a pair of factors that gives us negative 24. Out of these pairs, one of the numbers is negative, but when we combine, we get a positive 5. Let's run down the list. There's no way I could combine 1 and 24, whichever was negative, and get a positive 5, because if the 1 was negative and I combined it with 24, I'd get positive 23. If the 24 was negative, I'd get negative 23. There's no way. Okay, 2 and 12-- let's consider both cases. If 2 were negative, it'd be a positive 10. If 12 were negative, it'd be negative 10. That's not the pair. I've got a good feeling about this pair. If 3 were negative combined with a positive 8, I get positive 5; I would get it. Negative 3 plus 8 is positive 5. This is the pair that I want. I replace that trinomial in my numerator and write it by its factors. So, x minus 3 and x plus 8, and then I still have my denominator, x plus 8. Okay? I move my scratch work so it doesn't get confusing. Now you start crossing stuff out. So, that x plus 8, that's common-- I can cancel that out. And I'm left with x minus 3. And you're all done. Okay? All right. I hope you're feeling good about dividing polynomials and how to use the laws of exponents in certain problems and factoring in other problems to get through how to divide polynomials. Hope to see you here soon. Bye.

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In this program, students learn how to divide polynomials. In some cases, the division is really just a form of simplification. Sometimes, the problem requires long polynomial division. Part of the "Welcome to Algebra I" series.

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Runtime: 31 minutes

Welcome to Algebra I
Episode 1
31 minutes
Grade Level: 7 - 12
Welcome to Algebra I
Episode 2
25 minutes
Grade Level: 7 - 12
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Episode 3
18 minutes
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Episode 4
17 minutes
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Episode 5
22 minutes
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Episode 6
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Episode 7
24 minutes
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Episode 8
15 minutes
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Episode 9
25 minutes
Grade Level: 7 - 12
Welcome to Algebra I
Episode 10
16 minutes
Grade Level: 7 - 12