How come in partial fractions if you're given
You know that it can be rewritten:
But if you have
It can be rewritten as:
You've put two of the factors under C, granted they're both the same one (but I don't know what the significance of that is, it's probably the reason why it happens). Why stop there? Why don't we have a fourth fraction, call it D and shove the entire denominator under there? Also why don't we do this for questions that don't have a repeated root? Like my first example why don't we make a third, call it C and put the entire denominator under it?

Retsek
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 10032018 23:30

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 10032018 23:40
(Original post by Retsek)
How come in partial fractions if you're given
You know that it can be rewritten:
But if you have
It can be rewritten as:
You've put two of the factors under C, granted they're both the same one (but I don't know what the significance of that is, it's probably the reason why it happens). Why stop there? Why don't we have a fourth fraction, call it D and shove the entire denominator under there? Also why don't we do this for questions that don't have a repeated root? Like my first example why don't we make a third, call it C and put the entire denominator under it?
If we have then we can rewrite this as and and the key thing here is that the numerator is at most of order one less than the denominator. Further, we can take the second fraction and say:
hence why have multiple fractions with multiplicities of the repeated factor, though we skip this decomposition usually and just stick a C on the third fraction as it's all just constants mixing in so it doesnt matter.
We cannot have a fourth fraction because we run out of factors and their multiplicities.Last edited by RDKGames; 10032018 at 23:41. 
Retsek
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 10032018 23:55
(Original post by RDKGames)
In partial factors, we wish to split one fraction into several others and we know we can do so because when we come to add them into one, we must multiply all the denominators by each other to obtain a common one.
If we have then we can rewrite this as and and the key thing here is that the numerator is at most of order one less than the denominator. Further, we can take the second fraction and say:
hence why have multiple fractions with multiplicities of the repeated factor, though we skip this decomposition usually and just stick a C on the third fraction as it's all just constants mixing in so it doesnt matter.
We cannot have a fourth fraction because we run out of factors and their multiplicities.
It can be written as:
And then it gets pretty messy from there no? 
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 11032018 00:01
(Original post by Retsek)
How come in partial fractions if you're given
You know that it can be rewritten:
But if you have
It can be rewritten as:
You've put two of the factors under C, granted they're both the same one (but I don't know what the significance of that is, it's probably the reason why it happens). Why stop there? Why don't we have a fourth fraction, call it D and shove the entire denominator under there? Also why don't we do this for questions that don't have a repeated root? Like my first example why don't we make a third, call it C and put the entire denominator under it?
and what's the point of having three fractions with constant numerators.
The reason is the same reason why partial fraction decomposition is done at all : partial fractions are useful for integration. If you have
then you have three terms which are easy to integrate. 
RDKGames
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 11032018 00:06
(Original post by Retsek)
Okay brilliant explanation, so for this equation:
It can be written as:
And then it gets pretty messy from there no?
In fact, following on from this, I may as well tell you this:
When you got where represent the orders of polynomials and respectively, first check whether .
If so, the fraction can be reduced to where are some polynomials we wish to find and now we have . ie a fraction where the numerator has a lesser order than the denominator, and THEN you want to see whether factorises. If it does, then you split this fraction into its partial fractions as you've been doing.
This is what I applied to your question. I looked at the orders, they're the same, therefore there will be some polynomial of 0th order at the start, this is just a constant so let's call it A. Then you end up with some polynomial over which is fine, but this factorises into so we can split our fraction into two others where the numerators are at most of one degree less than these linear factors, ie the numerators are constants.
Hence we obtained the form I said at the start.
Anyway, I'm heading off now so if anything is unclear I can clarify tomorrow.Last edited by RDKGames; 11032018 at 00:16. 
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 11032018 00:25
(Original post by RDKGames)
It can be rewritten as . The A at the start becomes from the fact that the numerator and denominator have the same order, so division must yield a quotient that is some constant.
In fact, following on from this, I may as well tell you this:
When you got where represent the orders of polynomials and respectively, first check whether .
If so, the fraction can be reduced to where now we have . ie a fraction where the numerator has a lesser order than the denominator, and THEN you want to see whether factorises. If it does, then you split this fraction into its partial fractions as you've been doing.
The lowest order of the partial fraction will be 1, it explicitly does not have a constant for example in
Instead it can be represented by
Which goes to
Is this right? I think I get it nowLast edited by Retsek; 11032018 at 00:28. 
RDKGames
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 11032018 00:31
(Original post by Retsek)
Okay I see, so if we had
The lowest order of the partial fraction will be 1, it explicitly does not have a constant for example in
Instead it can be represented by
Which goes to
Is this right? I think I get it now
Your example of is an example of going by my notation. 
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 11032018 00:36
(Original post by RDKGames)
The example and its working is correct, though I'm not sure what you're referring to by "the lower order of the partial fraction will be 1" and the comment before that.
Your example of is an example of going by my notation.
Small side question, so all quotients with a denominator made up of linear factors (does count as a linear factor, if not I'm including those as well) can be written as a sum of quotients with constants for numerators (ie the backbone of partial fractions). What's the proof of this? 
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 11032018 00:42
(Original post by RDKGames)
It can be rewritten as . The A at the start becomes from the fact that the numerator and denominator have the same order, so division must yield a quotient that is some constant.
In fact, following on from this, I may as well tell you this:
When you got where represent the orders of polynomials and respectively, first check whether .
If so, the fraction can be reduced to where are some polynomials we wish to find and now we have . ie a fraction where the numerator has a lesser order than the denominator, and THEN you want to see whether factorises. If it does, then you split this fraction into its partial fractions as you've been doing.
This is what I applied to your question. I looked at the orders, they're the same, therefore there will be some polynomial of 0th order at the start, this is just a constant so let's call it A. Then you end up with some polynomial over which is fine, but this factorises into so we can split our fraction into two others where the numerators are at most of one degree less than these linear factors, ie the numerators are constants.
Hence we obtained the form I said at the start.
Anyway, I'm heading off now so if anything is unclear I can clarify tomorrow.
If the biggest power of x on top is same as, or bigger than the biggest power of x on the bottom, do the division first.
In the original question dividing 4+9x^2 by 49x^2 gives 1, remainder 18
so the original fraction can be written as 1 + 18/(32x)(3+2x)
and the second part of this (the fraction) can be written as 3/(32x) + 3/(3+2x) (your partial fractions bit)
hth 
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 11032018 00:57
(Original post by Retsek)
Oh shoot yeah your right.
Small side question, so all quotients with a denominator made up of linear factors (does count as a linear factor, if not I'm including those as well) can be written as a sum of quotients with constants for numerators (ie the backbone of partial fractions). What's the proof of this?
You dont always get constants in the numerators. For example;
goes to one fraction with a constant on top, and another with a linear function on top because isnt factorisable without getting into
You can google the proofs and such if you want as they can be founs on the internet I’m sure 
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 11032018 00:58
(Original post by RDKGames)
(x+1)^2 counts as a single linear factor of multiplicity 2.
You dont always get constants in the numerators. For example;
goes to one fraction with a constant on top, and another with a linear function on top because isnt factorisable without getting into
You can google the proofs and such if you want as they can be founs on the internet I’m sure
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