How do I find $\frac {x^3}{x(x-3)}$ partial fractions?

Question

How do I find $\frac {x^3}{x(x-3)}$ partial fractions? And in general, when the degree of the numerator is higher than the denominator's?

Thanks in advance for your assistance!

Answer 1

Before you can do what I would call partial fractions you have to have the degree of the numerator strictly smaller than the degree of the denominator. You get to that point by dividing the polynomials, getting a quotient and remainder (that thing you learned about in algebra but forgot).

It's been pointed out that the example you gave is a little trivial. A better example might be $\frac{x^3}{(x-1)(x-2)}$. That's the same as $\frac{x^3}{x^2-3x+2}$. You do the division and you get $$\frac{x^3}{x^2-3x+2} =x+\frac{3x^2-2x}{x^2-3x+2}=x+3+\frac{7x-6}{x^2-3x+2}$$(or seomthing like that). Now you handle the $x+3$ separately and do partial fractions on the last fraction.

Answer 2

First, we can simplify this expression as follows:

\begin{align*} \frac{x^3}{x(x-3)}&=\frac{x^2}{x-3}\;\,\qquad\quad\qquad\qquad\text{for }x\neq 0, 3\\ &=x+3+\frac{9}{x-3}\qquad\qquad\text{performing division of polynomials} \end{align*}