1. ## 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 $Bx + C$ ?

2. Because $x^2+1$ is irreducible over $\Bbb R$.

3. Originally Posted by lilaziz1
$\frac{2x-3}{x(x^2+1)} \Rightarrow 2x-3 = \frac{A}{x} + \frac{Bx+C}{x^2+1}$ 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.

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Partial-Fraction Decomposition: Repeated and Irreducible Factors

4. Originally Posted by lilaziz1
$\frac{2x-3}{x(x^2+1)} \Rightarrow 2x-3 = \frac{A}{x} + \frac{Bx+C}{x^2+1}$ 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 $C=2,\ A=-3$