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Math Help - Partial Fractions

  1. #1
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    Partial Fractions

     \frac{2x-3}{x(x^2+1)} \Rightarrow 2x-3 = \frac{A}{x} + \frac{Bx+C}{x^2+1} The question is, how does the numerator on  x^2 + 1 equal  Bx + C ?
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  2. #2
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    Because x^2+1 is irreducible over \Bbb R.
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  3. #3
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    Quote Originally Posted by lilaziz1 View Post
     \frac{2x-3}{x(x^2+1)} \Rightarrow 2x-3 = \frac{A}{x} + \frac{Bx+C}{x^2+1} The question is, how does the numerator on  x^2 + 1 equal  Bx + C ?
    If the denominator of the rational expression has an irreducible quadratic factor, then you have to account for the possible "size" of the numerator. If the denominator contains a degree-two factor, then the numerator might not be just a constant; it might be of degree one. So you would deal with a quadratic factor in the denominator by including a linear expression in the numerator.


    quoted from this site ...

    Partial-Fraction Decomposition: Repeated and Irreducible Factors
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    Quote Originally Posted by lilaziz1 View Post
     \frac{2x-3}{x(x^2+1)} \Rightarrow 2x-3 = \frac{A}{x} + \frac{Bx+C}{x^2+1} The question is, how does the numerator on  x^2 + 1 equal  Bx + C ?
    You need to combine the two fractions such that there will be no x^2 term in the resulting numerator

    \frac{A}{x}+\frac{Bx+C}{x^2+1}=\frac{x^2+1}{x^2+1}  \ \frac{A}{x}+\frac{x}{x}\ \frac{Bx+C}{x^2+1}

    =\frac{Ax^2+A+Bx^2+Cx}{x\left(x^2+1\right)}

    This allows the x^2 terms to cancel as B=-A,

    Also C=2,\ A=-3
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