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Partial fractions

  1. Jan 25, 2005 #1
    Given

    [tex]\frac{2+5x+15x^2}{\left (2-x\right )\left (1+2x^2\right )}=\frac{8}{2-x} + \frac{x-3}{1+2x^2}
    [/tex]

    I am asked to deduce the partial fractions of:

    [tex]\frac{1+5x+30x^2}{\left (1-x\right )\left (1+8x^2\right )}[/tex]

    I can solve it using my usual method, but that's not what the question requires. Any help? I can't see the link between the two.
     
  2. jcsd
  3. Jan 25, 2005 #2
    what is your USUAL method, and can people help you without knowing how UNUSUAL is not your USUAL?

    [tex]\frac{1+5x+30x^2}{\left (1-x\right )\left (1+8x^2\right )}=\frac{A}{1-x} + \frac{B}{1+8x^2}[/tex]

    and solve for A and B... this is the standard approach, is my method UNUSUAL? :surprised
     
  4. Jan 25, 2005 #3
    i have a not so usual method... substitude x=2u in your first equation... and you will get the second one. is this what you looking for?
     
  5. Jan 25, 2005 #4
    Yes, that's my usual way of solving such questions. But I still don't get it...
     
  6. Jan 25, 2005 #5

    This is wrong
    it should be :
    [tex]\frac{1+5x+30x^2}{\left (1-x\right )\left (1+8x^2\right )}=\frac{A}{1-x} + \frac{Bx +C}{1+8x^2}[/tex]

    Now determin A,B and C by adding up these fractions and then compare the numerator with the given fraction...

    marlon
     
    Last edited: Jan 25, 2005
  7. Jan 25, 2005 #6
    Marlon: thanks for your help, but the question specified that I deduce and not solve.

    Vincent: substituting x=2u works. Thanks.
     
  8. Jan 25, 2005 #7
    i know, but that answer is already given. I just wanted to point out that the given partial fractions were wrong,...that's all

    marlon
     
  9. Jan 25, 2005 #8
    me wrong?
    did I say B is constant?
     
  10. Jan 25, 2005 #9

    dextercioby

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    Compute A and B with your formula...

    Daniel.
     
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