# Partial fractions (with repeated terms in the denominator)Watch

#1
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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3 years ago
#2
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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3 years ago
#3
(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

from when you first learnt about partial fractions.

Therefore,

Now decompose the 2nd fraction on the right using

so that

and now you have an expansion of the original fraction into 3 separate terms.
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3 years ago
#4
(Original post by davros)
...
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 (and then form a continuous fraction)
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3 years ago
#5
(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 (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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3 years ago
#6
(Original post by davros)
...
Thanks!
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#7
(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

from when you first learnt about partial fractions.

Therefore,

Now decompose the 2nd fraction on the right using

so that

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