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Thread: Partial Fractions

  1. #1
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    Partial Fractions

    $\displaystyle \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 $\displaystyle x^2 + 1 $ equal $\displaystyle Bx + C $ ?
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    Because $\displaystyle x^2+1$ is irreducible over $\displaystyle \Bbb R$.
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  3. #3
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    Quote Originally Posted by lilaziz1 View Post
    $\displaystyle \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 $\displaystyle x^2 + 1 $ equal $\displaystyle 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.


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    Partial-Fraction Decomposition: Repeated and Irreducible Factors
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    Quote Originally Posted by lilaziz1 View Post
    $\displaystyle \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 $\displaystyle x^2 + 1 $ equal $\displaystyle Bx + C $ ?
    You need to combine the two fractions such that there will be no $\displaystyle x^2$ term in the resulting numerator

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

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

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

    Also $\displaystyle C=2,\ A=-3$
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