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# Partial fraction decomposition w/ repeated linear factors watch

1. Hey!

I'm an S6 in Scotland currently studying Advanced HIgher Maths. In class, we have just gone over Type II partial fractions (with repeated linear factors in the denominator). Our teacher told us that when you decompose this fraction, it will appear in the following form:

eg. x2 + 6x - 3
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2. (Original post by Sensei_Stig)
Hey!

I'm an S6 in Scotland currently studying Advanced HIgher Maths. In class, we have just gone over Type II partial fractions (with repeated linear factors in the denominator). Our teacher told us that when you decompose this fraction, it will appear in the following form:

eg. x2 + 6x - 3
---------------
You seem to have pressed the 'submit post' button before you finished typing, but anyway, with a repeated linear factor, you have to have a separate partial fraction for each power of the linear factor (in its denominator), up to the power present in the original rational expression - this is easiest to explain with an example:
(3x^2+6x-5)/(x-1)^2(2x-5) splits as A/x-1 + B/(x-1)^2 + C/2x-5
(4x^2+7x-9)/(3x-7)^3(x-2) splits as A/3x-7 + B/(3x-7)^2 + C/(3x-7)^3 + D/x-2
etc.
3. Oh sorry, I actually started typing this out and quit, it must have posted for me

I understand how to do it - and I apparently never got this typed out - but what I don't understand is why having one of each power (eg. 3x-7 AND (3x-7)2 AND (3x-7)3). Surely these would all combine on the denominator and leave a far higher power?
4. The reason for doing this is to make it easier to differentiate or integrate functions. You are trying to write the function in terms of a numeral divided by a variable or a numeral divided by a variable to a power (eg squared, cubed). For example, where A, B, and C are numbers, you would try to get your equation into the format of A/(3x-7) + B/(3x-7)^2 + C/(3x-7)^3. You would then say let (3x-7) = z then the equation becomes A/z + B/z^2 + C/z^3 which is much easier to integrate or differentiate.

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Updated: June 16, 2016
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