Partial Fractions

Rational Functions
We want to simplify a Rational Functions as much as possible

We can first long divide if the degree of numerator >= degree of denom
then factor the bottom
Then split into multiple fractions
for example

\begin{aligned} \frac{x^{3} + 2}{x^{3} - x} & = 1 + \frac{x + 2}{x^{3} - x} \\ \frac{x + 2}{x^{3} - x} & = \frac{x + 2}{x (x - 1) (x + 1)} \\ = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x + 1} \end{aligned}

If you can't split it up fully and there's a quadratic at the bottom, you have a linear function on top ( $\frac{B x + C}{+ b x + c}$ )
You can actually get it down to lines and parabolas always
(fundamental theorem of algebra)

\begin{aligned} \frac{A (x^{2} - 1) + B (x^{2} + x) + C (x^{2} - x)}{x (x - 1) (x + 1)} & = \frac{x + 2}{x (x - 1) (x + 1)} \end{aligned}

Now solve for $A, B, C$ by setting $x$ to $0, 1, - 1$ (zeros of linear parts)
AND/OR you can compare the coefficients of degrees with the numerator
eg. there is no $x^{2}$ in $x + 2$ , so $A + B + C = 0$

Rules:

If $Q$ factors into a bunch of distinct linear factors, then you can do $\frac{P (x)}{Q (x)} = \frac{A_{1}}{x - a_{1}} + \dots + \frac{A_{n}}{x - a_{n}}$
If $Q$ has a quadratic factor, then you can decompose into $\frac{B x + C}{x^{2} + b x + c}$
If $(x - a)^{m}$ is a repeated linear factor of $Q$ , then $\frac{A_{1}}{x - a} + \frac{A_{2}}{(x - a)^{2}} + \dots + \frac{A_{m}}{(x - a)^{m}}$
- You do this because you want unique terms, while still being factors of $Q$
Similarly, if $(x^{2} + b x + c)^{k}$ is a repeated quadratic factor of $Q$ , then $\frac{B_{1} x + C_{1}}{x^{2} + b x + c} + \frac{B_{2} x + C_{2}}{(x^{2} + b x + c)^{2}} + \dots + \frac{B_{k} x + C_{k}}{(x^{2} + b x + c)^{2}}$
It is always possible to factor $Q (x)$ into a product of linear factors and/or quadratic factors
- Fundamental Theorem of Algebra