Algebra (all content)
Intro to polynomial synthetic division
Sal shows how synthetic division of polynomials works. Created by Sal Khan.
Want to join the conversation?
- Which method is better in dividing polynomials? Synthetic or Long Division.(14 votes)
- Personal I love Synthetic division, try practicing with for a few days to a week until you understand how it really works because on a test or exam is so fast to do polynomial division with Synthetic division. Only downside to Synthetic division is that sometimes people have no idea what's going on it's almost like voodoo.(53 votes)
- is it necessary to know synthetic division for exams? (I do the IB)(12 votes)
- You don't really need to know synthetic division, or if you do, you should know long division too. It can save you valuable minutes, though.
Synthetic division reminds me of a FOIL - a thing from algebra to remember on how to distribute something like: (x+4)(x-3). Problem with Synthetic division and FOIL, is that they work only in couple simple cases and not in complex situations.
Edit: Apparently, I was wrong to some extent. Synthetic division proves to be useful when factoring polynomials what have more than two roots, e.g. x^4+2x^3+x-1=0. I won't go into a detail, but in terms of speed when you need to check like 6 roots, you can easily check them in half the time, compared to a long division.(22 votes)
- but what if youre deciding if something like 2x-1 is a factor of a polynomial function? i know you have to make it into the (x-k) form but how??(6 votes)
- You divide out the coefficient of x, to get a divisor of the form (x-k); you can then use synthetic division to check if (x-k) is a factor of the polynomial.
Here 2x-1 = 2(x-1/2), so you can use synthetic division to divide (x-1/2) into the polynomial to see if it is a factor, which would be equivalent to (2x-1) being a factor.(10 votes)
- What if the DENOMINATOR is not in 1st degree, but in 4th or 6th degree?(4 votes)
- My math teacher told me that this method can only be used for a linear denominator, such as x-k. So you can't use this for anything but a first degree denominator(6 votes)
- At3:27in this video I think I see a problem. Sal says that you multiply the 30 times a negative four when there is no negative four on the screen! Is this just me? Or if not, how did Sal get the negative four?(1 vote)
- The negative accounts for the subtraction involved in long division. He makes it all clear in the 'How it Works' video:
- Is this the same as Synthetic Substitution? because my teacher is calling it that. It seems to be the same thing.(5 votes)
- What is the difference between synthetic division and just division of polynomials?(5 votes)
- synthetic division is a big short cut if the divisor is only an X +/- any number.(1 vote)
- can you not do synthetic division when the x is squared? how do you divide two functions when it is squared?(2 votes)
- Use polynomial long division if the divisor has x^2 or higher term.(4 votes)
- Can we use synthetic division to solve polynomial that has more than one variable in numerator?
example: 2x^3 + cx^2 + 2x + 2 divided by x-2(2 votes)
- this is if you have 1 variable
you need the remainder, the last value in the synthetic division ( for the question in the video it was -121) and then plug it in the original equation with x=2 and y= the remainder.
You could work backward using synthetic division but its harder especially if you are new at this but to do that way you also need the remainder
if you have two variable you need 1 extra x and y value
if you have three variable you need 2 extra x and y value (3 x and y value in total)
and so on(3 votes)
- Not sure if someone already asked, but what if the 4 in the denominator was negative (x-4 instead of the example's x+4)? Would you bring it out as -4 or would you switch it to a positive 4?(2 votes)
- If the denominator was (x - 4), then your divisor for synthetic division becomes 4, not -4.(3 votes)
In this expression, we're dividing this third degree polynomial by this first degree polynomial. And we could simplify this by using traditional algebraic long division. But what we're going to cover in this video is a slightly different technique, and we call it synthetic division. And synthetic division is going to seem like a little bit of voodoo in the context of this video. In the next few videos we're going to think about why it actually makes sense, why you actually get the same result as traditional algebraic long division. My personal tastes are not to like synthetic division because it is very, very, very algorithmic. I prefer to do traditional algebraic long division. But I think you'll see that this has some advantages. It can be faster. And it also uses a lot less space on your paper. So let's actually perform this synthetic division. Let's actually simplify this expression. Before we start, there's two important things to keep in mind. We're doing, kind of, the most basic form of synthetic division. And to do this most basic algorithm, this most basic process, you have to look for two things in this bottom expression. The first is that it has to be a polynomial of degree 1. So you have just an x here. You don't have an x squared, an x to the third, an x to the fourth or anything like that. The other thing is, is that the coefficient here is a 1. There are ways to do it if the coefficient was different, but then our synthetic division, we'll have to add a little bit of bells and whistles to it. So in general, what I'm going to show you now will work if you have something of the form x plus or minus something else. So with that said, let's actually perform the synthetic division. So the first thing I'm going to do is write all the coefficients for this polynomial that's in the numerator. So let's write all of them. So we have a 3. We have a 4, that's a positive 4. We have a negative 2. And a negative 1. And you'll see different people draw different types of signs here depending on how they're doing synthetic division. But this is the most traditional. And you want to leave some space right here for another row of numbers. So that's why I've gone all the way down here. And then we look at the denominator. And in particular, we're going to look whatever x plus or minus is, right over here. So we'll look at, right over here, we have a positive 4. Instead of writing a positive 4, we write the negative of that. So we write the negative, which would be negative 4. And now we are all set up, and we are ready to perform our synthetic division. And it's going to seem like voodoo. In future videos, we'll explain why this works. So first, this first coefficient, we literally just bring it straight down. And so you put the 3 there. Then you multiply what you have here times the negative 4. So you multiply it times the negative 4. 3 times negative 4 is negative 12. Then you add the 4 to the negative 12. 4 plus negative 12 is negative 8. Then you multiply negative 8 times the negative 4. I think you see the pattern. Negative 8 times negative 4 is positive 32. Now we add negative 2 plus positive 32. That gives us positive 30. Then you multiply the positive 30 times the negative 4. And that gives you negative 120. And then you add the negative 1 plus the negative 120. And you end up with a negative 121. Now the last thing you do is say, well, I have one term here. And in this plain, vanilla, simple version of synthetic division, we're only dealing, actually, when you have x plus or minus something. So you're only going to have one term there. So you separate out one term from the right, just like that. And we essentially have our answer, even though it seems like voodoo. So to simplify this, you get, and you could have a drum roll right over here, this right over here, it's going to be a constant term. You could think of it as a degree 0 term. This is going to be an x term. And this is going to be an x squared term. You can kind of just build up from here, saying this first one is going to be a constant. Then this is going to be an x term, then an x squared. If we had more you'd have an x to the third, an x to the fourth, so on and so forth. So this is going to be equal to 3x squared minus 8x plus 30. And this right over here you can view as the remainder, so minus 121 over the x plus 4. This didn't divide perfectly. So over the x plus 4. Another way you could have done it, you could have said, this is the remainder. So I'm going to have a negative 121 over x plus 4. And this is going to be plus 30 minus 8x plus 3x squared. So hopefully that makes some sense. I'll do another example in the next video. And then we'll think about why this actually works.