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Homework Help: Alternative efficient ways to do partial fraction?

  1. Sep 28, 2005 #1
    what are alternative ways to do partial fraction?
    other than the normal method of mulplying the whole term over and comparing coefficient which is super headache method

    i heard of "cover-up method", gets the answer real quick

    anyone have details to this?
     
  2. jcsd
  3. Sep 28, 2005 #2

    TD

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    Homework Helper

    By choosing "smart zeroes" of the denominator, you can speed up the process but this won't always work (fully).

    For example, we wish to expand:

    [tex]\frac{{x + 1}}
    {{x^3 - 7x + 3}}[/tex]

    So you start by factoring and introducing the unknown coëfficiënts:

    [tex]\frac{{x + 1}}
    {{x^3 - 7x + 3}} = \frac{{x + 1}}
    {{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x + 3} \right)}} = \frac{A}
    {{x - 1}} + \frac{B}
    {{x - 2}} + \frac{C}
    {{x + 3}} = \frac{{A\left( {x - 2} \right)\left( {x + 3} \right) + B\left( {x - 1} \right)\left( {x + 3} \right) + C\left( {x - 1} \right)\left( {x - 2} \right)}}
    {{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x + 3} \right)}}[/tex]

    Then you set up the equation for the numerators to be the same:

    [tex]A\left( {x - 2} \right)\left( {x + 3} \right) + B\left( {x - 1} \right)\left( {x + 3} \right) + C\left( {x - 1} \right)\left( {x - 2} \right) = x + 1[/tex]

    Traditionally, you would now expand the entire left side and then group in powers of x to get a system, which is still doable.

    Another way to go on now would be to choose some values for x, preferably values which may simplify things, i.e. zeroes of the denominator:

    [tex]\begin{gathered}
    x = 1 \Rightarrow A\left( {1 - 2} \right)\left( {1 + 3} \right) = 2 \Leftrightarrow A = - \frac{1}
    {2} \hfill \\
    x = 2 \Rightarrow B\left( {2 - 1} \right)\left( {2 + 3} \right) = 3 \Leftrightarrow B = \frac{3}
    {5} \hfill \\
    x = - 3 \Rightarrow C\left( { - 3 - 1} \right)\left( { - 3 - 2} \right) = - 2 \Leftrightarrow C = - \frac{1}
    {{10}} \hfill \\
    \end{gathered} [/tex]

    Which gives the result:

    [tex]\frac{{x + 1}}
    {{x^3 - 7x + 3}} = \frac{{ - 1}}
    {{2\left( {x - 1} \right)}} + \frac{3}
    {{5\left( {x - 2} \right)}} - \frac{1}
    {{10\left( {x + 3} \right)}}[/tex]
     
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