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### Course: Algebra (all content) > Unit 10

Lesson 22: Synthetic division of polynomials# Why synthetic division works

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

## Want to join the conversation?

- 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?(96 votes)
- I would just divide everything by the coefficient, so that you get x + a while maintaining the equation.(77 votes)

- Why does algebraic long division work?(39 votes)
- 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)

- Will there ever be any skills practice concerning the division of polynomials?(23 votes)
- What if the answer doesn't have remainder?(5 votes)
- If it doesn't have a remainder, this is a beautiful thing! Just as when you divide 18 by 3 you get a nice "6" as opposed to 19 divided by 3. You still get the "6" but you have a little left over...1/3.(26 votes)

- is there a video about the fundamental theorem of algebra?(6 votes)
- There is a set of videos on the Origin and History of Algebra here:

http://www.khanacademy.org/math/algebra/introduction-to-algebra/overview_hist_alg/v/origins-of-algebra(16 votes)

- 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)
- 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)

i.e.

(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)

- If synthetic division works for polynomials, would it also work for integers?(4 votes)
- 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)

- Would synthetic division work with numbers like 2x or x^2? Or would it work only with x?(3 votes)
- no it only works with X. the coefficient and exponent must both be implied 1's. this is because, like sal just explained, since you can asume that your dividing by an X you can simplify the problem by only multiplying 1 term.(4 votes)

- When do you use synthetic division?(2 votes)
- You can also use it to find the answer for f(x). You can divide the coefficients of your polynomial (as Sal showed you in the video) by the x value you want to plug into the function. The remainder of the synthetic division will be f(x).(4 votes)

- 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)
- The divisor is the denominator. The divisor for the example in the video is x+4, so that's the denominator. The remainder will have a sign. If the remainder is -11, then it becomes "- the fraction 11/(x+4)".

Hope this helps.(3 votes)

## 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.