# How does polynomial long division work? Watch

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I'm Year 12 right now, going through C2 in Maths (Edexcel). The teacher has introduced polynomial long division and I understand how to do it and can confidently do it right. However, it does not make sense to me at all.

For example, (x^2 + x + 1 )/ (x + 1).

We take the x^2 and divide it by x to find the partial quotient. However, doesn't that mean we're just dividing the dividend by x in this case? I'm really confused as to how this actually works. It makes no sense!!

For example, (x^2 + x + 1 )/ (x + 1).

We take the x^2 and divide it by x to find the partial quotient. However, doesn't that mean we're just dividing the dividend by x in this case? I'm really confused as to how this actually works. It makes no sense!!

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I'm Year 12 right now, going through C2 in Maths (Edexcel). The teacher has introduced polynomial long division and I understand how to do it and can confidently do it right. However, it does not make sense to me at all.

For example, (x^2 + x + 1 )/ (x + 1).

We take the x^2 and divide it by x to find the partial quotient. However, doesn't that mean we're just dividing the dividend by x in this case? I'm really confused as to how this actually works. It makes no sense!!

**TKLovesScience**)I'm Year 12 right now, going through C2 in Maths (Edexcel). The teacher has introduced polynomial long division and I understand how to do it and can confidently do it right. However, it does not make sense to me at all.

For example, (x^2 + x + 1 )/ (x + 1).

We take the x^2 and divide it by x to find the partial quotient. However, doesn't that mean we're just dividing the dividend by x in this case? I'm really confused as to how this actually works. It makes no sense!!

e.g. with normal division, suppose I want to divide 1001 by 11.

I can do that the normal long division way (to get 91).

But I could also go: I don't feel like dividing by 11, so I'll just divide by 10 instead as it's easier. 1001/10 = 100.1, I'll just ignore the .1, so my first estimate is 100.

Then I subtract off, to find 1001 - 100 x 11 = -99. The key point here is that although 100 is a rubbish estimate, as long as I multiply that estimate by the actual divisor, everything I write is still valid.

And then at that point I note -99 = -9 x 11, so 1001 = 100 x 11 - 9 x 11 = 91.

In fact, if you have a polynomial division, and you replace x by a really big power of 10. (e.g. replace x by 1000000000), you'll find that (in general), the easiest way of doing the long division will be identical to the way you do polynomial division.

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This is what I did on paint.

You essentially do it like normal long division. You progressively remove factors, starting with the highest order term of x. In your example, you are left with a remainder of 1 and this is equivalent to like the remainder of 4/3 being equivalent to ( )

You essentially do it like normal long division. You progressively remove factors, starting with the highest order term of x. In your example, you are left with a remainder of 1 and this is equivalent to like the remainder of 4/3 being equivalent to ( )

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**TKLovesScience**)

I'm Year 12 right now, going through C2 in Maths (Edexcel). The teacher has introduced polynomial long division and I understand how to do it and can confidently do it right. However, it does not make sense to me at all.

For example, (x^2 + x + 1 )/ (x + 1).

We take the x^2 and divide it by x to find the partial quotient. However, doesn't that mean we're just dividing the dividend by x in this case? I'm really confused as to how this actually works. It makes no sense!!

The Euclidean division algorithm for integers says essentially that we can always chop them up like so, for e.g. for dividing 32 by 5:

32 = 6 x 5 + 2

i.e. we can find a quotient (6) and a remainder (2), with the remainder less than the divisor (here 5).

You can do a similar thing for polynomials, and the proof of it does it by matching up the leading terms only, at each stage, where after you have completed a stage, you subtract to find how big the "error polynomial" is that you have left to share out.

Since this approach works in the proof that division can be done in general for polynomials, it works just fine when you do it for real, as in your example. It looks somewhat weird but it does work, provably.

You can also do the same thing with numbers, so rather than using long division on say, 131/11, you can use the polynomial approach on , which is similar to the polynomial division of

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(Original post by

No, because we don't actually divide the dividend by x, we've just made an "estimate" of what the quotient will be, and we're going to subtract off that estimate times (x+1). The thing that makes the division process divide by a particular number is the subtracting off, not the estimate.

e.g. with normal division, suppose I want to divide 1001 by 11.

I can do that the normal long division way (to get 91).

But I could also go: I don't feel like dividing by 11, so I'll just divide by 10 instead as it's easier. 1001/10 = 100.1, I'll just ignore the .1, so my first estimate is 100.

Then I subtract off, to find 1001 - 100 x 11 = -99. The key point here is that although 100 is a rubbish estimate, as long as I multiply that estimate by the actual divisor, everything I write is still valid.

And then at that point I note -99 = -9 x 11, so 1001 = 100 x 11 - 9 x 11 = 91.

In fact, if you have a polynomial division, and you replace x by a really big power of 10. (e.g. replace x by 1000000000), you'll find that (in general), the easiest way of doing the long division will be identical to the way you do polynomial division.

**DFranklin**)No, because we don't actually divide the dividend by x, we've just made an "estimate" of what the quotient will be, and we're going to subtract off that estimate times (x+1). The thing that makes the division process divide by a particular number is the subtracting off, not the estimate.

e.g. with normal division, suppose I want to divide 1001 by 11.

I can do that the normal long division way (to get 91).

But I could also go: I don't feel like dividing by 11, so I'll just divide by 10 instead as it's easier. 1001/10 = 100.1, I'll just ignore the .1, so my first estimate is 100.

Then I subtract off, to find 1001 - 100 x 11 = -99. The key point here is that although 100 is a rubbish estimate, as long as I multiply that estimate by the actual divisor, everything I write is still valid.

And then at that point I note -99 = -9 x 11, so 1001 = 100 x 11 - 9 x 11 = 91.

In fact, if you have a polynomial division, and you replace x by a really big power of 10. (e.g. replace x by 1000000000), you'll find that (in general), the easiest way of doing the long division will be identical to the way you do polynomial division.

As a side question about an entirely different topic, would you happen to be able to explain why in statistics, to find the median of a discrete set of data we use (n+1)/2 but to find the median for continuous or grouped discrete data we use n/2?

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(Original post by

This is what I did on paint.

You essentially do it like normal long division. You progressively remove factors, starting with the highest order term of x. In your example, you are left with a remainder of 1 and this is equivalent to like the remainder of 4/3 being equivalent to ( )

**X_IDE_sidf**)This is what I did on paint.

You essentially do it like normal long division. You progressively remove factors, starting with the highest order term of x. In your example, you are left with a remainder of 1 and this is equivalent to like the remainder of 4/3 being equivalent to ( )

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reply

(Original post by

This is a pretty good question. I'm not sure that I can answer it fully, but I think that this approach works because it essentially copies the reasoning of the proof that there is a Euclidean division algorithm for polynomials.

The Euclidean division algorithm for integers says essentially that we can always chop them up like so, for e.g. for dividing 32 by 5:

32 = 6 x 5 + 2

i.e. we can find a quotient (6) and a remainder (2), with the remainder less than the divisor (here 5).

You can do a similar thing for polynomials, and the proof of it does it by matching up the leading terms only, at each stage, where after you have completed a stage, you subtract to find how big the "error polynomial" is that you have left to share out.

Since this approach works in the proof that division can be done in general for polynomials, it works just fine when you do it for real, as in your example. It looks somewhat weird but it does work, provably.

You can also do the same thing with numbers, so rather than using long division on say, 131/11, you can use the polynomial approach on , which is similar to the polynomial division of

**atsruser**)This is a pretty good question. I'm not sure that I can answer it fully, but I think that this approach works because it essentially copies the reasoning of the proof that there is a Euclidean division algorithm for polynomials.

The Euclidean division algorithm for integers says essentially that we can always chop them up like so, for e.g. for dividing 32 by 5:

32 = 6 x 5 + 2

i.e. we can find a quotient (6) and a remainder (2), with the remainder less than the divisor (here 5).

You can do a similar thing for polynomials, and the proof of it does it by matching up the leading terms only, at each stage, where after you have completed a stage, you subtract to find how big the "error polynomial" is that you have left to share out.

Since this approach works in the proof that division can be done in general for polynomials, it works just fine when you do it for real, as in your example. It looks somewhat weird but it does work, provably.

You can also do the same thing with numbers, so rather than using long division on say, 131/11, you can use the polynomial approach on , which is similar to the polynomial division of

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