# Partial Fractions

#### lilaziz1

$$\displaystyle \frac{2x-3}{x(x^2+1)} \Rightarrow 2x-3 = \frac{A}{x} + \frac{Bx+C}{x^2+1}$$ The question is, how does the numerator on $$\displaystyle x^2 + 1$$ equal $$\displaystyle Bx + C$$ ?

#### ojones

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

TheCoffeeMachine

#### skeeter

MHF Helper
$$\displaystyle \frac{2x-3}{x(x^2+1)} \Rightarrow 2x-3 = \frac{A}{x} + \frac{Bx+C}{x^2+1}$$ 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

$$\displaystyle \frac{2x-3}{x(x^2+1)} \Rightarrow 2x-3 = \frac{A}{x} + \frac{Bx+C}{x^2+1}$$ 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 $$\displaystyle C=2,\ A=-3$$