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Math Help - Partial fractions

  1. #1
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    Partial fractions

    The question has a linear factor in the denominator to the power of 3. I have only encountered 2 before. Have I started this question correctly? If I compare coefficients of x cubed the I find that A=2.
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    Re: Partial fractions

    the method used in your picture is not correct, when you take x=-3, the denominator will become 0.
    So you better to use comparing the coefficient when you encounter this kind of problem.
    Last edited by piscoau; September 14th 2011 at 08:33 AM.
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    Re: Partial fractions

    Quote Originally Posted by Stuck Man View Post
    The question has a linear factor in the denominator to the power of 3. I have only encountered 2 before. Have I started this question correctly? If I compare coefficients of x cubed the I find that A=2.
    You should actually use \displaystyle \frac{2x^3 - 17x^2 + 26x - 56}{(x - 3)^3(x + 1)} \equiv \frac{A}{x + 1} + \frac{B}{x - 3} + \frac{C}{(x - 3)^2} + \frac{D}{(x - 3)^3}
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    Re: Partial fractions

    Quote Originally Posted by piscoau View Post
    the method used in your picture is not correct, when you take x=-3, the denominator will become 0.
    So you better to use comparing the coefficient when you encounter this kind of problem.
    No, what he is doing is perfectly valid. (Except, of course, for the missing \frac{1}{x+1} term.) Yes, the original form has a denominator of x+ 3 but after multiplying through by the least common denominator, he has a polynomial that is equal to that for all x other than those values that make the denominator 0. And since a polynomial is continuous, the value of the polynomial at 3 gives the correct value for the rational function as well.
    Last edited by HallsofIvy; September 15th 2011 at 03:37 AM.
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    Re: Partial fractions

    Quote Originally Posted by HallsofIvy View Post
    No, what he is doing is perfectly valid. (Except, of course, for the missing [tex]\frac{1}{x+1} term.) Yes, the original form has a denominator of x+ 3 but after multiplying through by the least common denominator, he has a polynomial that is equal to that for all x other than those values that make the denominator 0. And since a polynomial is continuous, the value of the polynomial at 3 gives the correct value for the rational function as well.
    AND the missing \displaystyle \frac{1}{x - 3} term. When the OP gets a common denominator, the OP will get \displaystyle \frac{\alpha x + \beta}{(x - 3)^3}, there needs to be a square term in the numerator as well...
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