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Partial fractions

  1. Aug 27, 2005 #1

    I got
    it = 1/(2x+1) + 4.75/(x-1) + -2/(x-1)^2 + 8.75(x^2+5)

    My working was several pages so Im not gonna post it. I was wondering if any of you know if that is right? Are there any geniuses on here who can do them in there head?
  2. jcsd
  3. Aug 27, 2005 #2


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    Hmmm, I don't think it's correct.
    You can check your answer by pluging x in and check if they are equal.
    x = 0: The first one gives: -4.
    While the second gives: 38.
    So it's wrong...
    Viet Dao,
  4. Aug 27, 2005 #3


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    Factoring the denominator should give:

    [tex]2x^5 - 3x^4 + 10x^3 - 14x^2 + 5 = \left( {x - 1} \right)^2 \cdot \left( {x^2 + 5} \right) \cdot \left( {2x + 1} \right)[/tex]

    So propose the partial fractions:

    [tex]\frac{A}{{\left( {x - 1} \right)^2 }} + \frac{B}{{x - 1}} + \frac{{Cx + D}}{{x^2 + 5}} + \frac{E}{{2x + 1}}[/tex]

    Now, rather than working it out and rearranging in powers of x to get a large 5x5 system, try choosing values of x that simplify the calculations, i.e. x's for which parts become zero (zero's of the denominators)
  5. Aug 28, 2005 #4


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    Looks like you factored the denominator correctly
    try the cover up shortcut, each highest order term can be found by substituting the a in to the function except for the factor that is zero. Lower order terms are found by subtracting off higher order terms. For terms of the form x^2+a^2 x^2 may be substituted for to avoid complex numbers, or they may be treated with the others using complex numbers.
    factor denominator
    =1/(2x+1) [you were right]
    x=1 (order 2)
    =-2/(x-1)^2 [you were right]
    x=1 (order 1)
    =3/(x-1) [you were wrong]
    where the quotient without remainder is used
    here to make things simple we substitute for a^2 instead of a. We multiply numerator and denominator by the conjugate of the denominator to rationalize. The general form that results is (ua+v)/(x^2-a^2) we replace a with x in the answer.
    -x/(x^2+5) [you were wrong]
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