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Partial Fraction Decomposition

  1. Jul 3, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the Partial Fraction Decomposition.

    [tex]\frac{4x^2+2x-1}{x^2(x+1)}[/tex]

    2. Relevant equations



    3. The attempt at a solution

    [tex]\frac{4x^2+2x-1}{x^2(x+1)}=\frac{a}{x}+\frac{b}{x^2}+\frac{c}{(x+1)}[/tex]

    [tex]4x^2+2x-1=x^2(x+1)a+x(x+1)b+x^2(x)c[/tex]

    So i can solve for c by plugging in -1 for x but I'm not sure how to solve for a and b.
     
  2. jcsd
  3. Jul 3, 2010 #2

    vela

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    You have an extra factor of x on the right side. Fixing that may help you see how to solve for b.
     
  4. Jul 3, 2010 #3
    Extra factor? do you mean the right side has too many x? Are you saying that I multiplied by the LCD wrong?
     
  5. Jul 3, 2010 #4

    vela

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    Yes, when you multiplied by the denominator, you did it incorrectly.
     
  6. Jul 3, 2010 #5
    How is that wrong? You need to get an LCD of [tex]x^3(x+1)[/tex]

    So the only way to do that is to multiply a by [tex]\frac{x^2(x+1)}{x^2(x+1)}[/tex]

    b by [tex]\frac{x(x+1)}{x(x+1)}[/tex]

    and c by [tex]\frac{x^2(x)}{x^2(x)}[/tex]
     
  7. Jul 3, 2010 #6

    Hurkyl

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    That's not the LCD. Now, it is still a CD so it's fine to multiply by that -- but you need to multiply both sides by the same thing when solving an equation!

    P.S. it's just a system of equations.
    P.P.S. why not evaluate at 2 or 17 or anything else?
     
  8. Jul 3, 2010 #7

    vela

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    You may find it easier to think of it as multiplying both sides by the denominator of the LHS:

    [tex]x^2(x+1) \left[\frac{4x^2+2x-1}{x^2(x+1)}\right] = x^2(x+1)\left[\frac{a}{x}+\frac{b}{x^2}+\frac{c}{(x+1)}\right][/tex]

    When you simplify, you'll get

    [tex]4x^2+2x-1 = x(x+1)a + (x+1)b + x^2 c[/tex]
     
  9. Jul 3, 2010 #8
    Haha! your right that isn't the LCD.

    ok so now I have:

    [tex]4x^2+2x-1=x(x+1)a+(x+1)b+x^2c[/tex]

    ok so I found c=1 and b=-1 then I can set up a system of equations and find that A=3

    Thanks!
     
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