Get ready for Precalculus
- Intro to long division of polynomials
- Dividing quadratics by linear expressions (no remainders)
- Divide quadratics by linear expressions (no remainders)
- Dividing quadratics by linear expressions with remainders
- Dividing quadratics by linear expressions with remainders: missing x-term
- Divide quadratics by linear expressions (with remainders)
Intro to long division of polynomials
Any quotient of polynomials a(x)/b(x) can be written as q(x)+r(x)/b(x), where the degree of r(x) is less than the degree of b(x). For example, (x²-3x+5)/(x-1) can be written as x-2+3/(x-1). This latter form can be more useful for many problems that involve polynomials. The most common method for finding how to rewrite quotients like that is *polynomial long division*. Created by Sal Khan and CK-12 Foundation.
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- What about Synthetic Division? Is it something that recently came out? It is supposed to be shorthand version of Polynomial Division, where you just use the coefficient of each term and work it out that way. Look forward to your response!(211 votes)
- Sal has now made some video about synthetic division: http://www.khanacademy.org/math/algebra/polynomials/v/synthetic-division(213 votes)
- why is it that x goes 1 time in x square?(12 votes)
- if you don't see a number with an exponent, then it means the number's exponent is 1.
i.e 2 . you don't see any exponents but it has exponentof 1 because 2 is just 2..(20 votes)
- At2:40, how come Sal can just ignore the 1 in the divisor, and only worry about the highest degree term? Shouldn't he be dividing x^2 with the whole divisor x+1?
I don't see how dividing x^2 only by the largest degree term x, and then multiplying the result into the divisor is the same thing as dividing x^2 with the entirety of (x+1).(16 votes)
- Someone else already asked this, here's the answer they got :)
"This solution will become crystal clear when you start dividing by higher polynomials.
Consider long division using the following notation:
17568 = 1*10^4 + 7*10*^3 + 5*10^2 + 6*10^1 + 8 & 10^0
Divide this by 202 which is 2*10^2 + 0*10^1 + 2
Take out the null coefficient and divide 17568 by 202 using powers of ten. You start by diving the largest power of ten into the largest power of ten and then multiplying everything by that number and subtracting. Try it, you'll like it."(6 votes)
- is sythentic division the same thing as dividing polynomials?(7 votes)
- It's a method of dividing a polynomial by a binomial so in some cases, yes(15 votes)
- Why should we only consider the largest power when dividing? For example at3:03, he divides x^2 by x. Surely x^2/(x+1) is different from x^2/(x)(11 votes)
- 37 times 2x would be 74x, right?(0 votes)
- Yes! When multiplying a constant by a variable term, multiply the constant by the coefficient of the variable.(1 vote)
- I can't get my head around this for some reason, any tips?(5 votes)
- I couldn't either at first. I ended up watching the video a few more times and doing my own question from a textbook and then it clicked!(5 votes)
- At1:31, how does 2 go into 2x x times?(3 votes)
- He just means 2 times x equals 2x, just like 2 times 4 is 8, so just think of eight as 2x and 4 as x(6 votes)
- Why do you only consider the highest degree term in polynomial division it is not enough that it works please help as this is a fundamental part of mathematics which needs to be understood throughly and not just memorised(4 votes)
- It's just like long division of numbers.
When you divide 21 | 714 , you start with 7/2, not 7/21 or 7/2 and 7/1. In other words, you're sorting it into chunks biggest chunk first, then the next smaller, then the next smaller, and so on.(2 votes)
- is there such thing as a quad polynomial?(2 votes)
- A polynomial can have infinite length.(4 votes)
In this video, we're going to learn to divide polynomials, and sometimes this is called algebraic long division. But you'll see what I'm talking about when we do a few examples. Let's say I just want to divide 2x plus 4 and divide it by 2. We're not really changing the value. We're just changing how we're going to express the value. So we already know how to simplify this. We've done this in the past. We could divide the numerator and the denominator by 2, and then this would be equal to what? This would be equal to x plus 2-- let me write it this way-- it would be equal to, if you divide this by 2, it becomes an x. You divide the 4 by 2, it becomes a 2. If you divide the 2 by 2, you get a 1. So this is equal to x plus 2, which is pretty straightforward, I think. The other way is you could have factored a 2 out of here, and then those would have canceled out. But I'll also show you how to do it using algebraic long division, which is a bit of overkill for this problem. But I just want to show you that it's not fundamentally anything new. It's just a different way of doing things, but it's useful for more complicated problems. So you could have also written this as 2 goes into 2x plus 4 how many times? And you would perform this the same way you would do traditional long division. You'd say 2-- you always start with the highest degree term. 2 goes into the highest degree term. You would ignore the 4. 2 goes into 2x how many times? Well, it goes into 2x x times and you put the x in the x place. x times 2 is 2x. And just like traditional long division, you now subtract. So 2x plus 4 minus 2x is what? It's 4, right? And then 2 goes into 4 how many times? It goes into it two times, a positive two times. Put that in the constants place. 2 times 2 is 4. You subtract, remainder 0. So this might seem overkill for what was probably a problem that you already knew how to do and do it in a few steps. We're now going to see that this is a very generalizable process. You can do this really for any degree polynomial dividing into any other degree polynomial. Let me show you what I'm talking about. So let's say we wanted to divide x plus 1 into x squared plus 3x plus 6. So what do we do here? So you look at the highest degree term here, which is an x, and you look at the highest degree term here, which is an x squared. So you can ignore everything else. And that really simplifies the process. You say x goes into x squared how many times? Well, x squared divided by x is just x, right? x goes into x squared x times. You put it in the x place. This is the x place right here or the x to the first power place. So x times x plus 1 is what? x times x is x squared. x times 1 is x, so it's x squared plus x. And just like we did over here, we now subtract. And what do we get? x squared plus 3x plus 6 minus x squared-- let me be very careful-- this is minus x squared plus x. I want to make sure that negative sign only-- it applies to this whole thing. So x squared minus x squared, those cancel out. 3x, this is going to be a minus x. Let me put that sign there. So this is minus x squared minus x, just to be clear. We're subtracting the whole thing. 3x minus x is 2x. And then you bring down the 6, or 6 minus 0 is nothing. So 2x plus 6. Now, you look at the highest degree term, an x and a 2x. How many times does x go into 2x? It goes into it two times. 2 times x is 2x. 2 times 1 is 2. So we get 2 times x plus 1 is 2x plus 2. But we're going to want to subtract this from this up here, so we're going to subtract it. Instead of writing 2x plus 2, we could just write negative 2x minus 2 and then add them. These guys cancel out. 6 minus 2 is 4. And how many times does x go into 4? We could just say that's zero times, or we could say that 4 is the remainder. So if we wanted to rewrite x squared plus 3x plus 6 over x plus 1-- notice, this is the same thing as x squared plus 3x plus 6 divided by x plus 1, this thing divided by this, we can now say that this is equal to x plus 2. it is equal to x plus 2 plus the remainder divided by x plus 1 plus 4 over x plus 1. This right here and this right here are equivalent. And if you wanted to check that, if you wanted to go from this back to that, what you could do is multiply this by x plus 1 over x plus 1 and it add the two. So this is the same thing as x plus 2. And I'm just going to multiply that times x plus 1 over x plus 1. That's just multiplying it by 1. And then to that, add 4 over x plus 1. I did that so I have the same common denominator. And when you perform this addition right here, when you multiply these two binomials and then add the 4 up here, you should you get x squared plus 3x plus 6. Let's do another one of these. They're kind of fun. So let's say that we have-- we want to simplify x squared plus 5x plus 4 over x plus 4. So once again, we can do our algebraic long division. We can divide x plus 4 into x squared plus 5x plus 4. And once again, same exact process. Look at the highest degree terms in both of them. x goes into x squared how many times? It goes into it x times. Put it in the x place. This is our x place right here. X times x is x squared. x times 4 is 4x. And then, of course, we're going to want to subtract these from there. So let me just put a negative sign there. And then these cancel out. 5x minus 4x is x. 4 minus 0 is plus 4. x plus 4, and then you could even see this coming. You could say x plus 4 goes into x plus 4 obviously one time, or if you were not looking at the constant terms, you would completely just say, well, x goes into x how many times? Well, one time. Plus 1. 1 times x is x. 1 times 4 is 4. We're going to subtract them from up here, so it cancels out, so we have no remainder. So this right here simplifies to-- this is equal to x plus 1. And there's other ways you could have done this. We could have tried to factor this numerator. x squared plus 5x plus 4 over x plus 4. This is the same thing as what? We could have factored this numerator as x plus 4 times x plus 1. 4 times 1 is 4. 4 plus 1 is 5, all of that over x plus 4. That cancels out and you're left just with x plus 1. Either way would have worked, but the algebraic long division will always work, even if you can't cancel out factors like that, even if you did have a remainder. In this situation, you didn't. So this was equal to x plus 1. Let's do another one of these just to make sure that you really-- because this is actually a very, very useful skill to have in your toolkit. So let's say we have x squared-- let me just change it up. Let's say we had 2x squared-- I could really make these numbers up on the fly. 2x squared minus 20x plus 12 divided by-- actually, let's make it really interesting, just to show you that it'll always work. I want to go above quadratic. So let's say we have 3x to the third minus 2x squared plus 7x minus 4, and we want to divide that by x squared plus 1. I just made this up. But we can just do the algebraic long division to figure out what this is going to be or what this simplified will be. x squared plus 1 divided into this thing up here, 3x to the third minus 2x squared plus 7x minus 4. Once again, look at the highest degree term. x squared goes into 3x to the third how many times? Well, it's going to go into it 3x times. You multiply 3x times this, you get 3x to the third. So it's going to go into it 3x times. So you have to write the 3x over here in the x terms. So it's going to go into it 3x times, just like that. Now let's multiply. 3x times x squared is 3x to the third, right? 3x times squared plus 3x times 1. So we have a 3x over here. I'm making sure to put it in the x place. And we're going to want to subtract them. And what do we have? What do we have when we do that? These cancel out. We have a minus 2x squared. And then 7x minus-- because I just subtracted 0 from there-- 7x minus 3x is plus 4x, and we have a minus 4. Once again, look at the highest degree term. x squared and a negative 2x squared. So x squared goes into negative 2x squared negative 2 times. Negative 2, put it in the constants place. Negative 2 times x squared is negative 2x squared. Negative 2 times 1 is negative 2. Now, we're going to want to subtract these from there, so let's multiply them by negative 1, or those become a positive. These two guys cancel out. 4x minus 0 is-- let me switch colors-- 4x minus 0 is 4x. Negative 4 minus negative 2 or negative 4 plus 2 is equal to negative 2. And then x squared, now it has a higher degree than 4x, and the highest degree here, so we view this as the remainder. So this expression we could rewrite it as being equal to 3x minus 2-- that's the 3x minus 2-- plus our remainder 4x minus 2, all of that over x squared plus 1. Hopefully, you found that as fun as I did.