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    How come in partial fractions if you're given

    \dfrac{f(x)}{(x+1)(x+2)}

    You know that it can be rewritten:

    \dfrac{A}{x+1} + \dfrac{B}{x+2}

    But if you have

    \dfrac{f(x)}{(x+1)(x+2)^2}

    It can be rewritten as:

    \dfrac{A}{x+1} + \dfrac{B}{x+2} + \dfrac{C}{(x+2)^2}

    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?
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    (Original post by Retsek)
    How come in partial fractions if you're given

    \dfrac{f(x)}{(x+1)(x+2)}

    You know that it can be rewritten:

    \dfrac{A}{x+1} + \dfrac{B}{x+2}

    But if you have

    \dfrac{f(x)}{(x+1)(x+2)^2}

    It can be rewritten as:

    \dfrac{A}{x+1} + \dfrac{B}{x+2} + \dfrac{C}{(x+2)^2}

    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?
    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 \dfrac{x^2-4x+1}{(x+1)(x+2)^2} then we can rewrite this as \dfrac{A}{x+1} and \dfrac{Bx+C}{(x+2)^2} 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:

    \begin{aligned} \dfrac{Bx+C}{(x+2)^2} & = \frac{Bx+2B+C-2B}{(x+2)^2} \\ & = \frac{B}{x+2} + \frac{C-2B}{(x+2)^2} \\ & = \frac{B}{x+2} + \frac{D}{(x+2)^2}

    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.
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    (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 \dfrac{x^2-4x+1}{(x+1)(x+2)^2} then we can rewrite this as \dfrac{A}{x+1} and \dfrac{Bx+C}{(x+2)^2} 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:

    \begin{aligned} \dfrac{Bx+C}{(x+2)^2} & = \frac{Bx+2B+C-2B}{(x+2)^2} \\ & = \frac{B}{x+2} + \frac{C-2B}{(x+2)^2} \\ & = \frac{B}{x+2} + \frac{D}{(x+2)^2}

    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.
    Okay brilliant explanation, so for this equation:

    \dfrac{9+4x^2}{9-4x^2}

    It can be written as:

    \dfrac{A}{3+2x}+\dfrac{B}{3-2x}+\dfrac{Cx+D}{9-4x^2}

    And then it gets pretty messy from there no?
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    (Original post by Retsek)
    How come in partial fractions if you're given

    \dfrac{f(x)}{(x+1)(x+2)}

    You know that it can be rewritten:

    \dfrac{A}{x+1} + \dfrac{B}{x+2}

    But if you have

    \dfrac{f(x)}{(x+1)(x+2)^2}

    It can be rewritten as:

    \dfrac{A}{x+1} + \dfrac{B}{x+2} + \dfrac{C}{(x+2)^2}

    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?
    Just adding to what RDKGames said, you may be wondering why we don't just leave it in the form

    \dfrac{A}{x+1} + \dfrac{Bx+C}{(x+2)^2}

    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

    \dfrac{A}{x+1} + \dfrac{B}{x+2} + \dfrac{C}{(x+2)^2}

    then you have three terms which are easy to integrate.
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    (Original post by Retsek)
    Okay brilliant explanation, so for this equation:

    \dfrac{9+4x^2}{9-4x^2}

    It can be written as:

    \dfrac{A}{3+2x}+\dfrac{B}{3-2x}+\dfrac{Cx+D}{9-4x^2}

    And then it gets pretty messy from there no?
    It can be rewritten as A+\dfrac{B}{3+2x}+\dfrac{C}{3-2x}. 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 \dfrac{f_m(x)}{g_n(x)} where m,n represent the orders of polynomials f and g respectively, first check whether m\geq n.
    If so, the fraction can be reduced to h_{m-n}(x) + \dfrac{r_p(x)}{g_n(x)} where h,r are some polynomials we wish to find and now we have p<n. ie a fraction where the numerator has a lesser order than the denominator, and THEN you want to see whether g_n(x) 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 9-4x^2 which is fine, but this factorises into (3+2x)(3-2x) 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.
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    (Original post by RDKGames)
    It can be rewritten as A+\dfrac{B}{3+2x}+\dfrac{C}{3-2x}. 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 \dfrac{f_m(x)}{g_n(x)} where m,n represent the orders of polynomials f and g respectively, first check whether m\geq n.
    If so, the fraction can be reduced to h_{m-n}(x) + \dfrac{r_p(x)}{g_n(x)} where now we have p<n. ie a fraction where the numerator has a lesser order than the denominator, and THEN you want to see whether g_n(x) factorises. If it does, then you split this fraction into its partial fractions as you've been doing.
    Okay I see, so if we had \dfrac{f_2(x)}{g_3(x)}

    The lowest order of the partial fraction will be 1, it explicitly does not have a constant for example in \dfrac{x+1}{(x+3)^2}

    Instead it can be represented by \dfrac{Ax+3A+B-3A}{(x+3)^2}

    Which goes to \dfrac{A}{x+3} + \dfrac{C}{(x+3)^2}

    Is this right? I think I get it now
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    (Original post by Retsek)
    Okay I see, so if we had \dfrac{f_2(x)}{g_3(x)}

    The lowest order of the partial fraction will be 1, it explicitly does not have a constant for example in \dfrac{x+1}{(x+3)^2}

    Instead it can be represented by \dfrac{Ax+3A+B-3A}{(x+3)^2}

    Which goes to \dfrac{A}{x+3} + \dfrac{C}{(x+3)^2}

    Is this right? I think I get it now
    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 \dfrac{x+1}{(x+3)^2} is an example of \dfrac{f_1(x)}{g_2(x)} going by my notation.
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    (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 \dfrac{x+1}{(x+3)^2} is an example of \dfrac{f_1(x)}{g_2(x)} going by my notation.
    Oh shoot yeah your right.

    Small side question, so all quotients with a denominator made up of linear factors (does (x+1)^2 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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    (Original post by RDKGames)
    It can be rewritten as A+\dfrac{B}{3+2x}+\dfrac{C}{3-2x}. 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 \dfrac{f_m(x)}{g_n(x)} where m,n represent the orders of polynomials f and g respectively, first check whether m\geq n.
    If so, the fraction can be reduced to h_{m-n}(x) + \dfrac{r_p(x)}{g_n(x)} where h,r are some polynomials we wish to find and now we have p<n. ie a fraction where the numerator has a lesser order than the denominator, and THEN you want to see whether g_n(x) 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 9-4x^2 which is fine, but this factorises into (3+2x)(3-2x) 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.
    ... Another way of saying what RDK has just said:

    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 4-9x^2 gives -1, remainder 18

    so the original fraction can be written as -1 + 18/(3-2x)(3+2x)
    and the second part of this (the fraction) can be written as 3/(3-2x) + 3/(3+2x) (your partial fractions bit)

    hth
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    (Original post by Retsek)
    Oh shoot yeah your right.

    Small side question, so all quotients with a denominator made up of linear factors (does (x+1)^2 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?
    (x+1)^2 counts as a single linear factor of multiplicity 2.

    You dont always get constants in the numerators. For example;

    \dfrac{3x^2+3x+2}{(x+1)(x^2+1)} goes to one fraction with a constant on top, and another with a linear function on top because x^2+1 isnt factorisable without getting into \mathbb{C}

    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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    (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;

    \dfrac{3x^2+3x+2}{(x+1)(x^2+1)} goes to one fraction with a constant on top, and another with a linear function on top because x^2+1 isnt factorisable without getting into \mathbb{C}

    You can google the proofs and such if you want as they can be founs on the internet I’m sure
    Okay thanks for all your help!
 
 
 
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