# Welcome to Algebra I: Dividing Polynomials

31 minutes

(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) 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.

## Media Details

Runtime: 31 minutes