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Welcome to Algebra I: Adding and Subtracting Polynomials

46 minutes

Hi, guys, welcome to Algebra 1. Today's topic is going to focus on adding and subtracting polynomials. All that you know about simplifying expressions is really going to take you far on this topic. You ready to get started? Let's go. Let me take you back a bit to the distributive property. You're going to apply it when you're working with polynomials, when you add and subtract them, you apply this a bit.

(female describer) 2 times 4X plus one.

(instructor) Just to review, you remember the distributive property. With that 2 on the left, I can distribute it throughout what's in my parentheses. Somebody probably showed you these little arrows to show you what you were doing. So I can say, I need to multiply 2 times 4X, which is 8X, and I need to multiply 2 times positive 1, which is 2. You could say, "2 times that quantity, 4X plus 1." After I apply the distributive property, I could write that as 8X plus 2. Right? Let's apply that property a few more times, get it in our system. This expression: 7 plus 3 times the quantity X minus 2. For this one, I'm going to work down a bit. That might help this make a little more sense to you. Let's move that over slightly. Okay. So the 7 is just going along for the ride for a bit. Let me erase that dot so you don't think it's subtraction. 7 is just going along for the ride for a second, so 7 plus... I do need to apply my distributive property here, for the second bit. So 3 times X, that's 3X. And 3 times negative 2-- because that's how I interpret that, 3 times negative 2-- that's negative 6. So you'd write it as a minus 6. Then you look back at your problem and you can combine some like terms here. You do have your 3X, which has nothing else like it in that expression, so there's nothing to combine that with. So it's now just coming on down here. I have two constant terms that I can combine, to write it more simply. So 7 minus 6, okay-- so 7 minus 6, that's 1. So that means I can write that as a plus 1. So you apply the distributive property and you simplify that original expression to 3X plus 1. Okay? Let's keep going. This one, let's do the same thing. Let's move that over a little bit. So I have 5 minus that quantity (3X plus 4). That 5 is just going to go along for the ride for a minute. When you see something like this, I hope you remember, you have this minus and then this 3X plus 4 in parentheses. Treat that as if there is a 1 right in front of it. What's happening is you're distributing a negative 1 throughout what's in your parentheses. Okay? Negative 1 times 3X, that's a negative 3X, and then negative 1 times 4, that's a negative 4. Now I'm at the point where I have like terms to combine. This negative 3X doesn't have anything like it. It's the only term in this expression with an X, so I know that this is negative 3X. That's just going to bring on down. I do have this 5 minus 4. So 5 minus 4, I know that's 1, and it's a positive 1. So I can write that as plus 1. That original expression, it simplified to negative 3X plus 1. Jogged your memory a bit with the distributive property? Let's get into the polynomials. There are three steps to follow when you're adding and subtracting polynomials. You first want to rewrite the problem, then you want to combine like terms, then you want to make sure your final answer is in standard form. Remember, standard form, that's when you have your exponents where they go in descending order. So the leading term has the largest exponent and then you just go in descending order from there. Your final answer always needs to be in that form. So while we work with these problems, keep these three steps in mind. We'll rewrite it, combine like terms, then we'll make sure we're in standard form. Let's look at the first one. Deep breath, because I know it looks massive. There's a lot going on, but it's really going to break right on down to something that looks a lot more simple. Our first polynomial-- it's actually a trinomial, it's got three terms. So 7X cubed plus 6X squared plus 2x, and we're going to add that to 3X cubed minus 4X squared plus 7X minus 5. We're going to start by rewriting the problem. When you're adding polynomials, rewriting the problem is essentially just getting rid of the parentheses. When you're adding, the rewriting is just break those terms out of the parentheses. I have 7X cubed plus 6X squared plus 2X, plus 3X cubed minus 4X squared plus 7X minus 5. I just broke that problem out of the parentheses. Now what I'm going to do, just so this doesn't get too crowded here, is I'm going to get rid of that problem. Now we've broken it out of the parentheses, so let's just look at it by itself. I think it's going to help us look at this a little better. Let's click off of it so I can write. Here we go. Now we're at step two, where we need to combine like terms. How I find out which terms are alike is the terms have to have the same variable raised to the same power. The coefficient doesn't need to be the same, just the same variable raised to the same power. Those are our like terms. And since we know that we need our answer to be in standard form, let's start out with the largest exponent, so that by the end, we've got it written in the right form. Scan your terms and you notice that you have a 7X cubed and you have a 3X cubed. Let's take those two terms first. I've got the 7X cubed and I've got this 3X cubed. You notice how I took the sign out front, because it tells me what-- the sign of that term. I'm going to come off and do some scratch work over here. So 7X cubed plus 3X cubed. Now here, we're just adding like terms. We're not multiplying anything together, so we're not going to change the value of our exponents because we're not finding any products here; we're combining like terms. So 7X cubed plus 3X cubed, that's just 10X cubed. Essentially, all you're doing is adding your coefficients and just carrying along that variable to that power. Okay? I know we've got the first set handled, that 10X cubed, so that's out of the way. Go in descending order, so now let's look for exponents of two. And I notice I have this 6X squared and I have this negative 4X squared. All right? Let's combine those like terms. 6X squared-- and I'm combining it with a negative 4X squared, so I'm just going to treat that like subtraction. So 6X squared minus 4X squared. Remember, you're not messing with your exponents at all, you're really just focused on the coefficients. So 6X squared minus 4X squared-- well, 6 minus 4, that's just 2. So this is 2X squared. Okay? Now you've handled this. So you've handled the exponent of 3, you've handled the exponent of 2, so let's look for exponents of 1, which you really don't see. I'm going to start squaring things now, not as exponents, but just putting them in boxes to help me focus. I have a 2X squared and I have a positive 7-- I'm sorry, I have a 2X and I have a positive 7X, so I'm going to combine those like terms together. So 2X plus 7X. So, essentially, I'm just combining like terms here. So 2 plus 7, that's 9, so this is 9X. All right? I'll combine those: 2X plus 7X, that's 9X. That's taken care of. All that's left is that lonely little negative 5. That's our constant term; I'm just going to write it down here. Now we're at the point where we can write our answer and just make sure it's in standard form. I'm going to bring all these individual answers together. I'm going to write this as... Let's see the next thing to erase, make some room. Let's get rid of our problem because now we have the answer. And I'll write it right up here. Let's get that out of the way. When we added those polynomials together, it all really simplifies to 10X cubed... plus 2X squared... plus 9X minus 5. And that is our answer. That is what we get when we add those two polynomials together. And you can double check and we do have it in standard form. Our exponents are going from largest and then in descending order. Okay? You ready to try another one? Let's do it. All right. "10A cubed minus 3A squared plus 4," and we're going to add that to "negative 6A cubed plus A squared plus 1." Let's bring that up a little bit. You remember, step one is rewrite the problem. And when we're adding, essentially that's just breaking the terms out of the parentheses. So 10A cubed... minus 3A squared plus 4. So now look here. You see we have plus this negative 6A cubed? We don't like to have the signs bump into each other like that. Instead of writing "plus negative 6A cubed," I'm just going to write "minus 6A cubed." Didn't change the meaning at all because adding a negative means the same thing as subtraction. Remember that from Pre-Algebra. I can just rewrite plus negative 6A cubed as just minus 6A cubed. Then just break the other terms out of parentheses: A squared plus 1. Okay? Now we're at the point where we can combine like terms, so I'm going to give us a little more room up here. Let's get rid of our problem and let's bring this up. We can get some space to work. Now look for your like terms you can combine. Essentially, you're looking for the same variable raised to the same power, to identify your like terms. I see I have this 10A cubed and I have this negative 6A cubed, so I can combine this one and this one. I'll come off and do a little scratch work. So 10A cubed minus 6A cubed. You're just subtracting your coefficients here. So 10A cubed minus 6A cubed, that's 4A cubed, because 10 minus 6 is 4, right? All right, you handled that. Looking for your exponents in descending order, so you scan your terms and you see you have this negative 3A squared and this positive A squared, so those are your next two you want to combine. So negative 3A squared plus A squared-- you remember, when you don't see a coefficient, there's an invisible 1. So negative 3A squared plus 1A squared, that's just negative 2A squared. So you've handled those square terms. Now I see I have two constants left. I have this positive 4 and I have this positive 1. So 4 plus 1, that's 5. And you've combined all the terms you can combine, so you're ready to write your answer. I have some space, so I'll write it here. We're just bringing all those individual answers together, since we kind of worked out our problem in chunks. So 4A cubed minus 2A squared plus 5, because that 5 down there is positive. And you are all done because you have it in standard form. All right? Starting to feel a little better about it? Let's try another one. It usually takes three to feel like, "Okay, I think I got it, I think I'm ready to try one." Here we have 8B to the 4th minus 5B squared plus B, and we're adding that to 3B squared plus 5B. You remember step one? Rewrite the problem. And when we're adding, that's essentially just breaking it out of parentheses. So 8B to the 4th minus 5B squared plus B. We're going to add that to 3B squared plus 5B. All right. Now we're at step two, where we can combine like terms. This is where I get us some more space to work. Let's get that out of the way and bring that up a bit. Now let's get to work. I scan the problem and I know I want my answer in standard form, so I'm always going to look for the largest exponent first, to start identifying where my like terms are. I see this first term has an exponent of 4. It's the largest one, but I notice there are no other terms that have an exponent of 4. So there's nothing to combine that term with; it just keeps going along for the ride. 8B to the 4th just continued on being 8B to the 4th. It didn't combine with anything. Keep it in mind that we're trying to go in descending order with our exponents. We look for powers of 3. We don't see any powers of 3. We look for powers of 2, and I do see a couple terms that have a power of 2. I see this negative 5B squared and I see this positive 3B squared. I can combine those. Negative 5B squared plus 3B squared. Remember, essentially we're focusing on the coefficients, so negative 5 plus 3, that's negative 2. So this is negative 2B squared. All right. We've handled that. And I see all we're left with here is two terms that both have the variable B, and those are our last pair of like terms to combine. We have this b plus 5B. So B plus 5B. Remember, when you don't see a coefficient, there's an invisible 1. So 1B plus 5B, that is 6B. Look back at the polynomial, there's nothing else to combine. Now we're at the part where we can write our answer. I'll write it up at the top and get that out of our way. Then we can get the answer up here. All righty. Let's bring our individual answers all together, because we solved our problem in chunks. So 8B to the 4th minus 2B squared plus 6B. It's in standard form because the exponents are going in descending order. All done. All right? You ready to try one? Okay, how about two? Go ahead and press pause, take a few minutes. Take your time and work through these problems. Take it in chunks, just like we did. It'll be easier to manage that way. When you're ready to check your answers against mine, press play.

(female describer) Simplify. Number 1: Open parentheses, 11Y squared plus 3Y minus 2, close parentheses, plus, open parentheses, 4Y squared plus Y plus 5. Number 2: Open parentheses, negative 5X cubed plus 7X minus 1, close parentheses, plus, open parentheses, 8X cubed plus 3X squared plus 2X plus 2. All right, you ready to check? Let's check these. That first one: 11Y squared plus 3Y minus 2, plus 4Y squared plus Y plus 5. The answer there, you should've got 15Y squared plus 4Y minus 3. On the second one, we had negative 5X cubed plus 7X minus 1, and we were adding that to 8X cubed plus 3X squared plus 2X plus 2. And the answer, you should've got 3X cubed plus 3X squared plus 9X plus 1. If you want to see how I did that, keep on with me. Remember step one, when you're adding, essentially you're just breaking those terms out of the parentheses, is the first step. So 11Y squared plus 3Y minus 2, plus 4Y squared plus Y plus 5. Now that I've got it broken out of parentheses, give us some room to work. I can get rid of that and move this up a bit. All right. Now you're at the point where you need to scan your terms because you've got to find the like terms, you've got to combine. We know we want our answer to be in standard form, look for the terms that have the largest exponent first. I see my largest exponent in this polynomial is 2, and the terms that have that exponent are 11Y squared, and then this positive 4Y squared, so I do some scratch work. 11Y squared plus 4Y squared, that's 15Y squared. All right? So we've handled this. Now I see I have 3Y plus 1Y. Right? So 3Y plus 1Y, because there's an invisible 1 that I don't see. 3Y plus 1Y, that's 4Y. So I've handled those terms. And I have two constant terms left. I have this negative 2 and this positive 5. So negative 2 plus 5, that's a positive 3. Those are my last two terms that I needed to combine. Now you can write the answer. Since we worked the problem in chunks, we'll take all our chunks and put them together. So 15Y squared plus 4Y plus 3. That's how I got that first one. Need to see the next one? Okay. Going to bring that up a bit. Here we go. Negative 5X cubed plus 7X minus 1, plus 8X cubed plus 3X squared plus 2X plus 2. So remember for adding, your first step is to break it out of parentheses. So negative 5X cubed plus 7X minus 1, plus 8X cubed plus 3X squared plus 2X plus 2. So I've broken it out of parentheses, and just to help us focus, I'm going to get rid of that original problem and just zone in on this one. Now I'm looking for my like terms to combine, and because I know I want my answer in standard form, I'll start with my largest exponents first. I see that I have negative 5X cubed and this positive 8X cubed. So negative 5X cubed plus 8X cubed, that's 3X cubed. So you've handled those terms. Now I'm going in descending order, so I'm looking for X squared; there's only one little term that has an X squared. So it just comes along for the ride, there's nothing to combine it with. So I know 3X squared is just going to be around. Okay? Now I see I have 7X plus 2X. Right? So 7X-- oop, funky-looking X there-- plus 2X... that equals 9X. All right? So I handled that. And then I had two constant terms left. I have this negative 1 combined with this positive 2. So negative 1 plus 2, that's a positive 1. Now we have all our individual chunks and we just need to bring it together for our answer. I think I left enough space to write it here. So 3X cubed plus 3X squared... plus 9X plus 1. And that's how we got that answer. All right? Okay. So you've handled addition; now let's go to subtraction. Think back to one of those early problems that we did in this lesson, where we handled this problem with the distributive property. So to jog your memory, remember what we did on this one.

(describer) 5 minus, open parentheses, 3X plus 4, close parentheses.

(instructor) We said we have a five out front; it just stayed along for the ride. And we had to subtract that quantity in the parentheses, so we said it's minus this. There's a 1 here I don't see, so essentially I'm distributing this negative 1 throughout this quantity in here. So negative 1 times 3X, that's minus 3X. Negative 1 times 4, that was a negative 4. We said, "Okay, now we need to combine like terms." The negative 3X didn't have anything alike, so it just came along for the ride-- that term right there, we've handled it. Then we had this 5 minus 4, and 5 minus 4 is 1. So plus 1. So remember how we did this. The part that I want you to focus in on this, is that it was a subtraction type problem. It was 5 minus that 3X plus 4, and I want you to pay attention to what we had to do with distributing that negative 1. We use that same process when we subtract polynomials. Look at this one. We have 8X squared plus 5A, minus 4A squared plus 2A minus 6. Let's give ourselves a little room; I'll bring that up a bit. You remember when we're dealing with polynomials and we're adding and subtracting them, the first step is to rewrite the problem. Now, when we're adding, rewriting is essentially just breaking those terms out of the parentheses, but when we're subtracting, it's not just that. We have to distribute this negative 1, that's kind of invisible, that we're kind of not seeing. That's the extra step that's involved when you're subtracting polynomials. I look at that first term, or that first binomial-- that I'm going to break out of the parentheses; nothing's going on with that one. As far as this trinomial is concerned, this second batch, there is a minus 1 here that I don't see; it's kind of invisible. So I need to distribute this negative 1 to each of these terms before I can go any further with this problem. So negative 1 times 4A squared, that is-- let's show the work, just so it makes sense. So negative 1 times 4A squared, that's negative 4A squared. That first term is minus 4A squared. Now we need to multiply that negative 1 times the 2A, so negative 1 times 2A-- that's negative 2A, so minus 2A. Now I need to distribute that negative 1 to that negative 6, so negative 1 times negative 6, that's a positive 6, so plus 6. Now that I've rewritten the problem-- and because it was subtraction, it involved me distributing that negative 1 throughout-- now I can continue on to simplify this expression. Now it's pretty much the same as when we were adding. I think I left myself enough room to work here. I see I have this 8A squared minus 4A squared, all righty. So let me-- I know what I can do. I will get rid of this bit and show that work right there. So 8A squared minus 4A squared, that is 4A squared. Okay. So I've handled this. Now I have 5A-- and it's positive-- minus 2A, right? So 5A minus 2A, that's 3A. You've handled those terms. All you have left now is that constant term, that 6. It just goes along for the ride; you've got a 6. Now we're at the point, because we solved our problem in chunks, we'll bring our chunks together and then we'll have our answer. I would have 4A squared... plus 3A plus 6. And you are all done. Subtraction isn't too much unlike addition; you just have that extra step where you have to distribute that negative 1 throughout that second set of terms. Let's try another one. Okay, 7Y to the 4th plus 3Y squared minus 1, minus 6Y to the 4th plus 4Y squared minus 2Y plus 1. Step one: Rewrite the problem. And because we are subtracting, remember to use the distributive property for the second set. For the first set, just break it out of parentheses. So 7Y to the 4th plus 3Y squared minus 1. All right. Now that I'm at my second set, remember there is an invisible 1 here. There is a negative 1, because you're subtracting, that you have to distribute throughout those four terms. You have negative 1 times 6Y to the 4th. I'll come over here, so negative 1 times 6Y to the 4th, that's negative 6Y to the 4th. So it's minus 6Y to the 4th. And I have negative 1 times 4Y squared, so negative 1 times 4Y squared, that's negative 4Y squared, so minus 4Y squared. And I can see things are about to run into each other here, so let's just bring that down a little bit. Okay. Then I need to distribute this negative 1 to that negative 2Y. So negative 1 times negative 2Y, that's a positive 2Y. All right, so plus 2Y. Then I have this negative 1 times positive 1, all right? So negative 1 times positive 1, that's negative 1. I have a minus 1 right here, all right? Now we're at the point where we can combine like terms. I'm going to get that last bit of work out of our way because we're going to need some extra space to get a little scratch work in. All right, so let's see. I want to go in descending order of exponents, so which exponent is my largest in this case? I'm scanning the terms; it's 4. I have a 7Y to the 4th and I have a negative 6Y to the 4th. So 7Y to the 4th minus 6Y to the 4th... that is 1Y to the 4th. And I don't have to write that 1. So I've handled that. I don't have any X cubed-- I'm sorry, Y cubed terms, but I do have Y squared terms. I see I have a 3Y squared right here, and I have a negative 4Y squared there, so set's combine those: 3Y squared minus 4Y squared. So 3Y squared minus 4Y squared, that's negative 1Y squared. All right? So now I've handled those. And I see I do have a 2Y term, so that just is going along for the ride. I can squeeze it in right here. 2Y is just carrying along. And I do have a couple of constant terms here. I see I have this negative 1 and another negative 1. Let's get some things out of our way here. This problem got large. Let's get this out of the way. And let's do a little erasing and let's bring some stuff up a bit. Let's bring that up and let's bring the work up. Now we've got some more room for scratch work. All righty, so we were doing negative 1 minus 1. So negative 1 minus 1, that's negative 2. Now we have combined all the terms that we can, so we're at the point where we can write our final answer. Y to the 4th minus Y squared plus 2Y minus 2. It's in standard form; we are all done. All right? I think you've got one more in you. Let's try another one before you try them on your own. Negative 5X cubed minus X squared plus 3X, minus negative 4X cubed plus 4X squared. All right. Let's bring that up a bit. And let's get to work. So that first polynomial, essentially I'm breaking it out of parentheses. There's no multiplication going on out here that's telling me I need to do anything else. Negative 5X cubed minus X squared plus 3X, okay. Now I'm subtracting, so it's like I'm distributing a negative 1 throughout these terms. So I need negative 1 times negative 4X cubed-- negative 1 times negative 4X cubed. That's a positive 4X cubed, all righty. So plus 4X cubed. Then I have this negative 1 times 4X squared. So negative 1 times 4X squared, that is negative 4X squared, all righty? So minus 4X squared. We've gotten everything out of the parentheses, we did our distribution, so now we're at the point pf combining like terms. Let's get this old work out of the way, make room for some new scratch work. I know I want to go in descending order of exponents. My largest exponent is 3 in this problem. So I have a negative 5X cubed plus a 4X cubed. So negative 5X cubed plus 4X cubed, that's negative 1X cubed, right? Okay. So I've handled that. And I noticed I have some X squared terms. So negative X squared minus 4X squared. So negative X squared minus 4X squared, that's a negative 5X squared. You've handled the X squared terms, then all that's left is that 3X. There's nothing to combine with it, so it's just going along for the ride. Now we've handled everything we can as far as combining like terms, so we're at the point of writing our final answer. We would have negative X cubed minus 5X squared plus 3X. Got it in standard form, so we are all done. Okay? All right, why don't you try a couple? Go ahead and press pause and take a few minutes, then really take your time and work your way through these problems. To compare your answers with me, press play.

(describer) Simplify. Number 1: Open parentheses, negative 6C to the 5th plus 5C cubed plus C, close parentheses, minus, open parentheses, 4C to the 5th plus C cubed minus 8C. Number 2: Open parentheses, 5X cubed plus 6X squared minus 1, close parentheses, minus open parentheses, negative 2X cubed plus X squared plus 2. Ready to check? Let's go. For that first one, negative 6C to the 5th plus 5C cubed plus C, minus 4C to the 5th plus C cubed minus 8C. That one was negative 10C to the 5th plus 4C cubed plus 9C. That should've been your answer for the first one. 5X cubed plus 6X squared minus 1, minus negative 2X cubed plus X squared plus 2. That one, 7X cubed plus 5X squared minus 3. If you want to see how I did those... For the first trinomial, essentially we just need to get it out of the parentheses because there's no multiplication of anything here that I need to handle. So I would have negative 6C to the 5th plus 5C cubed plus C, okay? And because I'm subtracting, there's a negative 1 out here that I need to distribute throughout those terms. I need to do negative 1 times 4C to the 5th, negative 1 times C to the 3rd, negative 1 times negative 8C. I'll come off to the side and do that scratch work. So negative 1 times 4C to the 5th, that's negative 4C to the 5th, right? So minus 4C to the 5th. Then I have negative 1 times C to the 3rd. So negative 1 times C to the 3rd. That's negative C to the 3rd, right? So minus C to the 3rd. Then I have negative 1 times negative 8C. So negative 1 times negative 8C, and that's a positive 8C. That means I have a plus 8C. If you notice, when you're subtracting, essentially those terms right there, where I'm distributing that negative 1 throughout, their signs just become the opposite of whatever they were to start with. If you notice that pattern, that when you distribute the negative 1, it changes all the signs in that second group, you can jump to that step; that's fine. Algebra is all about patterns that lead to shortcuts. If you continue and do that distribution and really show the steps, that's fine too, because you'll get really comfortable with what we're doing here, with this process. Keep that in mind and let's keep going. We're at the point of combining like terms, so let me get us more room here to work. Let's get that out of the way. Let's erase this right here, let's get rid of that. Let's move this up. Now we've got some room to work here. So I know I want my answer in standard form, so scan your terms-- you're looking for the largest exponent first. And I see in this problem, my largest exponent's a 5. I'm going to combine negative 6C to the 5th with negative 4C to the 5th, right? So negative 6C to the 5th minus 4C to the 5th, that's negative 10C to the 5th. Right? Okay, so I've handled those. I don't have any C to the 4th terms, but I do have some C to the 3rd terms. So 5C to the 3rd, and I see I have this minus C to the 3rd. So 5C cubed minus C cubed, that's 4C cubed. Right? Okay. So we handled the C cubed terms. I see I have two terms left and they're alike-- they're both C. Or they both have a variable for C. I have this plus C and I have this plus 8C-- they're both positive-- so C plus 8C, that's 9C. All right? And I've handled all the terms in that polynomial. Now I can write my answer in standard form. All righty, so I'd have negative 10C to the 5th plus 4C to the 3rd plus 9C. And you are all done with that one. Okay? Let's look at how I did that next one. All right. For this one, let's move it up a bit. Okay. What you want to do first, check out that first polynomial. There's no multiplication or anything going on out front, so just break it out of parentheses. 5X cubed plus 6X squared minus 1. Now I'm subtracting, so remember you're distributing this negative 1 throughout these terms. I need to multiply each of those terms by negative 1. I'll come off to the side and do a little scratch work. For the first term, negative 1 times negative 2X cubed. So negative 1 times negative 2X cubed, that's a positive 2X cubed. Okay, so plus 2X cubed. Now I have this negative 1 times X squared. So negative 1 times X squared, that's a negative X squared. All righty. So minus X squared for that bit. Then I have a negative 1 times this positive 2. So negative 1 times 2, and that's negative 2. So, come up here, I've got my minus 2. Then we said you might have noticed it just basically changes all the signs in that second bit when you distribute that negative 1. Maybe you took that shortcut; maybe you did all the work. Either way is fine. Now we've reached the point of combining like terms, so let's give ourselves a little more room to work up here. Let's get rid of that, and let's get rid of this, and get that out of our way... and move this up. Okay, let's start combining. I know I want my answer in standard form, so I'm going to look for the largest exponent first. I see I have 5X cubed, 2X cubed, so I can combine those. So 5X cubed plus 2X cubed, that's 7X cubed. You've handled those two terms. Now you're scanning again. You see you have some X squared terms. I have 6X squared and minus X squared. So 6X squared minus X squared, that's 5X squared. You've handled the X squared terms. Then you look again and you have two constant terms left to combine. So I have this negative 1 and that negative 2. Negative 1 minus 2, that's negative 3. Now we can put each of our individual chunks together and write our final answer in standard form. So 7X cubed plus 5X squared minus 3, and you are all done. All right? Okay, well, I hope you really got a grip on how to add and subtract polynomials, and I hope you saw how the distributive property really helps you with those subtraction problems. See you soon for more Algebra 1. Bye, guys.

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In this program, students learn the steps for adding or subtracting polynomials. First, remove the parentheses surrounding the terms. Next, combine the like terms and solve the problem. Part of the "Welcome to Algebra I" series.

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Runtime: 46 minutes

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