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Math Help - Partial Fractions . . . HELP!!!

  1. #1
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    Question Partial Fractions . . . HELP!!!

    Find the partial fractions for the following:


    3x - 2x + 8
    ------------------------
    x + 2x + 4x + 8



    9x + 21x - 24
    --------------------------
    x + 4x - 11x - 30


    My biggest problem with these is that I have trouble factoring the bottom cubic in the second problem, but I'm having trouble with everything else also. Every time I try to solve the problems I end up with a different answer than the time before. I don't know what I am doing wrong!


    I shall be eternally grateful to whoever helps me solve these problems. So grateful, in fact, that I shall bestow upon them this bear hug.

    . . . Wow. It's obviously sleepy time for me. My true weirdness is oozing into your website. Terribly sorry about that.
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by potato92 View Post
    Find the partial fractions for the following:


    3x - 2x + 8
    ------------------------
    x + 2x + 4x + 8



    9x + 21x - 24
    --------------------------
    x + 4x - 11x - 30


    My biggest problem with these is that I have trouble factoring the bottom cubic in the second problem, but I'm having trouble with everything else also. Every time I try to solve the problems I end up with a different answer than the time before. I don't know what I am doing wrong!


    I shall be eternally grateful to whoever helps me solve these problems. So grateful, in fact, that I shall bestow upon them this bear hug.

    . . . Wow. It's obviously sleepy time for me. My true weirdness is oozing into your website. Terribly sorry about that.
    Question 2:

    Hint:

    \frac{9x^2+21x-24}{x^3+4x^2-11x-30} = \frac{{9x^2+21x-24}}{(x+2)(x+5)(x-3)}

    You need to find the partial fraction for the above term:

    Try:

     \frac{{9x^2+21x-24}}{(x+2)(x+5)(x-3)} = \frac{A}{x+2} + \frac{B}{x+5}+\frac{C}{x-3}

    solve this now. You should deninitely have partial fractions in your book/notes. question 1 would also be similar to this. if you cannot solve, show your work on where you get stuck!
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  3. #3
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    w00t!

    Thanks harish! I think those factors were what I needed.
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  4. #4
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    If your problem had a 1 in the numerator, there's a nice formula to bypass messy algebra which I'll demonstrate in this specific case:

    [1/(x + 2)(x + 5)(x - 3)] = A/(x + 2) + B/(x + 5) + C/(x - 3)

    A = 1/(5 - 2)\cdot(-3 - 2)
    B = 1/(2 - 5)\cdot(-3 - 5)
    C = 1/(2 - -3)\cdot(5 - -3)

    This formula works whenever you have a 1 in the numerator and linear factors in the denominator.
    Last edited by wonderboy1953; April 16th 2010 at 12:04 PM. Reason: grammar
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