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Math Help - Solving Partial Fractions

  1. #1
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    Solving Partial Fractions

    Q1)

     \frac{2x+4}{x^2 + 2x - 3}

    From my understanding, I have to always factorize the denominator of the fraction to one of these 3 formulas to proceed on:

    1. \frac{P(x)}{(ax+b)(cx+d)} = \frac{A}{ax+b} + \frac{B}{cx+d}

    2. \frac{P(x)}{(ax+b)(cx+d)^2} = \frac{A}{ax+b} + \frac{B}{cx+d} + \frac{C}{(cx+d)^2}

    3. \frac{P(x)}{(ax+b)(cx^2+dx+e)} = \frac{A}{ax+b} + \frac{Bx+C}{cx^2+dx+e} where cx^2+dx+e cannot be factored into linear factors (omg i haven't touch on this yet)

    However I tried and failed to factorize the denominator to substitute to any of the formulas above.


    Please enlighten me! I just need to change them into one of the stated formula and I can to solve it further by myself. Your every effort is appreciated.. Thank you.
    Last edited by transcendentx; August 23rd 2011 at 06:35 AM.
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  2. #2
    Super Member Quacky's Avatar
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    Re: Solving Partial Fractions

    We have:

    \frac{2x+4}{x^2+2x-3}

    Taking the discriminant of the quadratic on the denominator:

    2^2-4(1)(-3)

    =16

    This tells us that this quadratic will factor, as 16 is a perfect square (think of the quadratic formula, and what would happen if you substituted).

    Soon, we deduce that the factors are (x+3)(x-1) by trial and error (fortunately, -3 doesn't give many options).

    Can you take it from there?

    Edit: I'm sorry you had to wait so long for a response.
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  3. #3
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    Re: Solving Partial Fractions

    Thank youfor your help! No worries about the me taken.
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