Partial fractions (with repeated terms in the denominator) Watch

gloria97
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Why, when a fraction has repeated linear terms in its denominator e.g. (11x2+14x+5)/[(x+1)2(2x+1)] does it have to be split into three partial fractions, A/(x+1) + B/(x+1)2 + C/(2x+1)?
When my first saw this example, my initial reaction was to split it into A/(x+1)2 +B/(2x+1), but after working through this, I realised my method was wrong. Why doesn't it work? I don't want a worked answer to the example because I already know what it is. I just want a genuine logical reason to why the former works and the latter doesn't.
Thanks
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Hunkybm
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Yes it has to be split into A/(x+1) + B/(x+1)2 + C/(2x+1), if you think about it if you split it into only two fractions you're not really simplifying it down? You need to separate all the factors of the denominator and (x+1) is a factor! Remember, partial fractions are used to simplify. Hope this helped!
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davros
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(Original post by gloria97)
Why, when a fraction has repeated linear terms in its denominator e.g. (11x2+14x+5)/[(x+1)2(2x+1)] does it have to be split into three partial fractions, A/(x+1) + B/(x+1)2 + C/(2x+1)?
When my first saw this example, my initial reaction was to split it into A/(x+1)2 +B/(2x+1), but after working through this, I realised my method was wrong. Why doesn't it work? I don't want a worked answer to the example because I already know what it is. I just want a genuine logical reason to why the former works and the latter doesn't.
Thanks
I can never remember these things myself, but the good news is you can work it out for yourself from the basics.

You know that you can write

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

from when you first learnt about partial fractions.

Therefore,

\dfrac{1}{(x+1)^2(2x+1)}  = \dfrac{1}{x+1}[\dfrac{1}{(x+1)(2x+1)} ]

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

Now decompose the 2nd fraction on the right using

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

so that

\dfrac{B}{(x+1)(2x+1)} \equiv \dfrac{BA}{x+1} + \dfrac{B^2}{2x+1}

and now you have an expansion of the original fraction into 3 separate terms.
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MathMeister
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(Original post by davros)
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Say you disregarded the expression raised to the 1 but had the same expression raised to the 2 (and the other one) then wouldn't the numerator be Bx+C? Because for example  1/(x+1)= (x+1)/(x+1)^{2} (and then form a continuous fraction)
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davros
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(Original post by MathMeister)
Say you disregarded the expression raised to the 1 but had the same expression raised to the 2 (and the other one) then wouldn't the numerator be Bx+C? Because for example  1/(x+1)= (x+1)/(x+1)^{2} (and then form a continuous fraction)
Not quite sure what you're trying to say, but my brain's getting tired and I'm "Latexed out" with all these fractions

But if you're going where I think you're going, note that whenever you have something like
(Bx + C)/(x+1)^2 you can always rearrange the numerator to "look like" the linear factor in the denominator i.e. Bx + C = Bx + B - B + C = B(x+1) + C - B and then do a polynomial division to get constant/(x+1) + (another constant)/(x+1)^2.

I think the important thing is to be flexible enough to recognize that you can do lots of clever things with fractions like writing x/(x+1) = 1 - (1/(x+1)) when you're integrating for example, rather than trying to memorize every single combination that can come up
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MathMeister
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(Original post by davros)
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Thanks!
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gloria97
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(Original post by davros)
I can never remember these things myself, but the good news is you can work it out for yourself from the basics.

You know that you can write

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

from when you first learnt about partial fractions.

Therefore,

\dfrac{1}{(x+1)^2(2x+1)}  = \dfrac{1}{x+1}[\dfrac{1}{(x+1)(2x+1)} ]

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

Now decompose the 2nd fraction on the right using

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

so that

\dfrac{B}{(x+1)(2x+1)} \equiv \dfrac{BA}{x+1} + \dfrac{B^2}{2x+1}

and now you have an expansion of the original fraction into 3 separate terms.
Thanks sooooo much!!
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