Partial Fractions

Mar 2010
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14
\(\displaystyle \frac{2x-3}{x(x^2+1)} \Rightarrow 2x-3 = \frac{A}{x} + \frac{Bx+C}{x^2+1} \) The question is, how does the numerator on \(\displaystyle x^2 + 1 \) equal \(\displaystyle Bx + C \) ?
 
May 2010
274
67
Los Angeles, California
Because \(\displaystyle x^2+1\) is irreducible over \(\displaystyle \Bbb R\).
 
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skeeter

MHF Helper
Jun 2008
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\(\displaystyle \frac{2x-3}{x(x^2+1)} \Rightarrow 2x-3 = \frac{A}{x} + \frac{Bx+C}{x^2+1} \) 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
 
Dec 2009
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1,342
\(\displaystyle \frac{2x-3}{x(x^2+1)} \Rightarrow 2x-3 = \frac{A}{x} + \frac{Bx+C}{x^2+1} \) 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 \(\displaystyle C=2,\ A=-3\)