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Lyione · Oct 16, 2012, 01:34 PM

Please help A.S.A.P...
3x-1/(x^2-9)(x^2-1)

ebaines · Oct 16, 2012, 02:07 PM

Given:

$ \frac {3x-1}{(x^2-9)(x^2-1)} $

First step is to factor the denominator into it's lowest order constituents:

$ (x^2-9)(x^2-1) = (x+3)(x-3)(x+1)(x-1) $

Now set up the 4 fractions with unknown numerators:

$ \frac {3x-1}{(x^2-9)(x^2-1)} = \frac A {x+3} + \frac B {x-3} + \frac C {x+1} \frac D {x-1} $

Now it's a matter of finding values for A, B, C and D that satisy this. Can you take it from here?

Lyione · Oct 16, 2012, 02:20 PM

Thank you so much for your help! :)
Ummm... sorry but I am kind of slow in this...
Can you show one example to find A/B/C/D ?