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What special considerations do you need to take when decomposing the following fraction and why?

I'm trying to decompose the following:

$$ \frac{s^3-1}{(s^2+6)^2(s+12)^2} $$

$$ \frac{s^3-1}{(s^2+6)^2(s+12)^2} = \frac{A}{(s^2+6)^2} + \frac{B}{(s+12)^2} $$

It obviously isn't correct. What to do?

Nosrati
  • 29,995
Bob Shannon
  • 1,561

1 Answers1

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When doing Partial Fractions you need to consider all of the increasing powers and should have (see the handy table in the link):

$$ \dfrac{s^3-1}{(s^2+6)^2(s+12)^2} = \dfrac{As + B}{(s^2+6)} + \dfrac{Cs + D}{(s^2+6)^2}+ \dfrac{E}{(s+12)} + \dfrac{F}{(s+12)^2} $$

You should arrive at:

  • $A = \dfrac{758}{140625}$
  • $B = \dfrac{6841}{562500}$
  • $C = -\dfrac{67}{1875}$
  • $D = -\dfrac{167}{3750}$
  • $E = -\dfrac{758}{140625}$
  • $F = -\dfrac{1729}{22500}$
Amzoti
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  • Hmm, why do we use Ax+B form for $s^2+6$ an A form for $s+12$? Is it because $s^2+6$ is a quadratic term and $s+6$ is a linear term? – Bob Shannon Nov 18 '13 at 00:57
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    There is a set of rules and you can see a nice and handy table here http://tutorial.math.lamar.edu/Classes/Alg/PartialFractions.aspx. If you recall those four simple rules, they always work. – Amzoti Nov 18 '13 at 00:58
  • Might have to burn this textbook and replace it with Paul's notes instead. +1 – Bob Shannon Nov 18 '13 at 01:02
  • @Bob: It is always good to have more than one reference laying around! I would look for other books in the library to see if there are some that better suit your likes. Trust me, I have been there many a time! :-) – Amzoti Nov 18 '13 at 01:05