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Why synthetic division works

Sal explains why synthetic division gives you the same result as traditional algebraic long division. Created by Sal Khan.

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  • male robot hal style avatar for user Omkar A. Katta
    Sal said there is a way to use synthetic division even if there is a coefficient in front of x, but I haven't found any videos so far. How would you do it then? factor out the coefficient?
    (97 votes)
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  • aqualine ultimate style avatar for user Ben Lou (hai!)
    Why does algebraic long division work?
    (39 votes)
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    • piceratops ultimate style avatar for user Tony
      It is actually fairly simple, and just requires you to take a step back and ask yourself what the problem you are facing actually means instead of getting confused by all of the variables. It is the same basic principal of dividing whole numbers just taken to the next level. Basically what is happening is you are multiplying the denominator X times so that it is equal to the numerator. In normal division, 42/6 is saying "How many times do you multiply 6 to get 42?" Its the exact same concept. Except in this case it is slightly a bit abstract, it is "How many times do you multiply your denominator to equal your numerator?"

      These examples you would normally do by factoring out values, but just bear with me here and ill try to help you understand that they are actually the same thing and the same process. Starting out simple, if you have the expression 4x / 4. You don't think of it this way because your brain is doing it subconciously, you just realize that 4 over 4 is 1 with an x left over. You could write it as 1x, but the answer is just x. Now thinking in terms of what the "/" part of division(a fraction) actually means, "How many times do you multiply 4 to get "4x." The answer is "1x," which just simplifies to "x."

      Now lets add another term to the numerator, say you have the expression (8x + 8y) / 4 setup as a fraction. Your brain automatically goes, "Hey, 8x over 4 is just 2x." You can then rewrite the 8x/4 as just 2x, you simplified the fraction to a whole number. Lets not simplify the 8y just yet. You then end up to the expression "2 + 8y/4". In long division, 8y/4 is the remainder, what was left over in the original division you didn't do. If you set it up long division wise now as 8x + 8y divided by 4 you see this work itself out. 4 goes into 8x 2x times (4 * 2x = 8x.) You can then take the 8x out of the numerator as its equivalent whole number: 2x. You then look at what is remaining in the fraction (hence, the remainder) and say "How many times does my denominator go into this term?" 4 goes into 8y 2y times. Again, taking the 8y out of the numerator. If you are still lost, while looking at the fraction, always think of it like it is written in plain english... "How many times does my denominator go into my numerator?" or "How many times does 4 go into 8y?"

      This will keep going on and on as you add more terms. It is always the same question though, and you can always think of it as "How many times do I need to multiply the denominator so that I can subtract a term out of the numerator?" (starting with the highest degree term first.) Keep going until you can no longer remove any terms out of the numerator without adding any extra terms. I hope this helped, I watched Sal's videos too and learned how to do it very easily, and I too was curious as to exactly what was going on. You just really need to sit down and solve problems of increasing complexity. Start with a whole number division problem, ask yourself "Why am I doing this, and what is it doing?" Then setup numerator with a variable, and ask yourself that same question. It will always be the exact same reason no matter if you have whole numbers or a variable raised to the 100th power.
      (149 votes)
  • leaf green style avatar for user Harper Beeland
    Will there ever be any skills practice concerning the division of polynomials?
    (23 votes)
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  • male robot hal style avatar for user Brainstorm
    What if the answer doesn't have remainder?
    (5 votes)
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  • spunky sam blue style avatar for user Kevin Saenz
    is there a video about the fundamental theorem of algebra?
    (6 votes)
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  • male robot donald style avatar for user phillipboudreau5
    I think if there is a number in front of x you just factor it out of the denominator, and then ignore it and perform synthetic division normally. Then divide your answer by the number that you factored out.
    (2 votes)
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    • purple pi purple style avatar for user Stephanie
      Not quite:
      (2x^3+4x^2-5x+6)/(2x -3) = (1/2 (2x^3+4x^2-5x+6))/(1/2(2x -3)) is true.
      However, for remainder, you need to multiply 2 back.

      (2x^3+4x^2-5x+6)/(2x -3)
      = (x^2 + 7/2 x + 11/4) + (14 1/4)/ (2x -3) ----------(1)
      (1/2 (2x^3+4x^2-5x+6))/(1/2(2x -3))
      = (x^2 + 7/2 x + 11/4) + (7 1/8)/(1/2(2x -3)) ---(2)
      if we multiply (2x -3) on both side of (1) we get:
      (2x^3+4x^2-5x+6) = (2x -3)(x^2 + 7/2 x + 11/4) + (14 1/4)
      if we multiply (2x-3) on both side of (2), (1/2 cancel out) we get
      (2x^3+4x^2-5x+6) = (2x -3)(x^2 + 7/2 x + 11/4) + (7 1/8)/(1/2)
      (2x^3+4x^2-5x+6) = (2x -3)(x^2 + 7/2 x + 11/4) + (7 1/8)*2
      That is why you need to multiply 2 back on the remainder.
      (5 votes)
  • duskpin ultimate style avatar for user ‧͙⁺˚*・༓𝐸𝓁𝒾𝓏𝒶𝒷𝑒𝓉𝒽 𝒴.༓・*˚⁺‧͙
    If synthetic division works for polynomials, would it also work for integers?
    (4 votes)
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    • piceratops ultimate style avatar for user DarshK
      No synthetic division will not work for integers. You can test this by taking a three digit number, like 224, and dividing it(for this example 2). 200....20....4. Next you can divide it by the negative of 2, -2. If you follow the synthetic division process, you will not reach the correct conclusion. Thus synthetic division does not work for integers.
      (2 votes)
  • leaf blue style avatar for user larsandersland
    Would synthetic division work with numbers like 2x or x^2? Or would it work only with x?
    (3 votes)
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  • blobby green style avatar for user Jordynn Omura
    When do you use synthetic division?
    (2 votes)
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  • primosaur ultimate style avatar for user Milton
    the remainder has either a + or a - at the end of the result. how do you know which to use. example: ...x^3+...x^2+...x+...-11/x+4 or +11/x+4? or +11?x-4? or -11/x-4?
    (2 votes)
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Video transcript

What I want to do now is simplify the exact same expression but do it with traditional algebraic long division. And hopefully we'll see why synthetic division actually gives us the exact same result. We'll be able to see the connections between synthetic division and algebraic long division. So let's get started. So if we're doing algebraic long division and set it up right over here, the first thing we want to think about is how many times does the highest degree term here, which is our x, go into the highest degree term here, which is our 3x to the third power. Well x goes into 3x to the third power 3x squared times. So we'll write it in the x squared place, 3x squared times. Now you might already see your parallel. When we did the synthetic division, we dropped this 3 straight down, and this 3 represented 3x squared. So this 3 and this 3x squared are really representing the same thing. But you might be saying, well here we have to do some thinking. We had to say x goes into 3x to the third 3x squared times. Here, we just mindlessly dropped this 3 straight down. How did that work? The reason why we were able to mindlessly drop this 3 straight down is because we assumed, to do this basic type of synthetic division, that we had just an x right over here. We didn't have a 3x. We didn't have a 4x. We didn't have an x squared. We just had an x. And if you divide an x into whatever this highest degree term is, your coefficient is going to be the exact same thing. It's just going to be one degree lower. So you go from a 3x to the third to 3x squared-- so the exact same coefficient. And now, in our little synthetic division right over here, that right over there is the x squared term. So you went from 3x to the third to 3x squared. We essentially divided it by x. And we could blindly do it because we knew, we assumed, that we were dealing with just a 1x to the first right over here. But let's keep on going and see the parallels and see why we're essentially doing the exact same thing. Now let's take this 3x squared and multiply that times x plus 4. So 3x squared times x-- I'll do it in that white color-- it's going to be 3x to the third. And 3x squared times 4 is going to be 12x squared. And now we'll want to subtract this. So now we subtract. We subtract this out. These guys cancel out, and you have 4x squared minus 12x squared. And so that will give you negative 8x squared. So once again, you're probably seeing some parallels. You had the 4x squared over here. You have the 4x squared over there. We just wrote the coefficient, but that's what it represented. 4x squared, we wrote the 4 there. Then we essentially subtracted 12x squared. And the way we got that 12, we multiplied 3 times 4, and then we subtracted. Here we're multiplying 3 times negative 4. We're essentially multiplying 3 times 4 and then subtracting. That's why we put that negative there, so we don't have to keep remembering to subtract this row. So we could just keep adding them. But that's essentially what we did. We multiplied 3 times this 4. And now we subtract it. We get that negative 12x squared. And then we subtracted, and you got negative 8x squared. And you might say, oh, is this the same negative 8 as this right over here? Not quite yet, because over here, this negative 8 literally represents negative 8x. This is actually part of our simplification. When we divide this into that, we got 3x squared minus 8x plus 30. So over here in the algebraic long division, we then say, how many times does x plus 4 go into negative 8x squared? Well x goes into negative 8x squared negative 8x times. So this is the key right over here. x goes into it negative 8x times. And once again, the reason why we could just put this negative 8 here is we know that we are dividing just by a 1x. So you're going to have the exact same coefficient, just one degree lower. So this right over here is our x term, and you see it right over there, just like that. So a lot of the simplification just comes from the idea that we are assuming with the synthetic division that this is a 1x. But let's just keep going. So you have a negative 8x times this business right over here gives you negative 8x times x is negative 8x squared. And then you have negative 8x times 4, which is negative 32x. And we can bring down all of this business right over here, just so it becomes a little bit simpler. So you have a negative 2x. And then over here, you have a minus 1. And once again, when you're doing traditional algebraic long division, you're going to subtract this from that up there. So if we're going to subtract, that's like adding the negative. These characters cancel out. You have a negative 2x plus 32x. That gives us a positive 30x. And then we can bring down that negative 1 if we want. I'll do that yellow color, actually. We'll bring down that negative 1. So this 30 has the same coefficient here. But this 30 should be up here. This is going to be part of our final answer. And to get that, once again, it all comes from the fact that we know that we had an x here when we did the synthetic division. 30x divided by x is just going to be 30. That 30 and this 30 is the exact same thing. And then we multiply. 30 times x is 30x. Actually, let me write the 30x in that white color because that's the convention I've been using-- 30x. 30 times 4 is 120. And then we are going to subtract this from that. So we get negative 1 minus 120 is negative 121, which, right over there, is our remainder-- which is exactly what we got over there. So hopefully you see the connection. Because we are assuming that we are dividing by x plus or minus something, we were able to make some simplifying assumptions. Whenever you divide this by an x, you know it's going to have the same coefficient, just one degree lower. And we kept doing that. And so it allowed us to do it a little bit simpler, a little bit faster, and using less space.