Question

How do irreducible quadratic denominators complicate partial-fraction decomposition?

Answer 1

If you were trying to find the partial fraction decomposition of

`x/(x^2-9)`,

you would break up the denominator into `(x+3)(x-3)`, and the problem would be set up as follows:

`x/((x+3)(x-3))=A/(x+3)+B/(x-3)`

which would be simplified to be

`x=A(x-3)+B(x-3)`

which is very easily solved.

However, if the problem were

`x/(x^4-81)`,

the denominator would factor into `(x^2+9)(x^2-9)=(x^2+9)(x+3)(x-3)`.

The `x^2+9` term cannot be factored (over the real numbers) and is and thus is an irreducible quadratic.

When setting up the partial fraction decomposition for something like this, it looks like:

`x/((x^2+9)(x+3)(x-3))=(Ax+B)/(x^2+9)+C/(x+3)+D/(x-3)`

When continuing to solve this, the `Ax+B` term necessitated by an irreducible quadratic term will only complicate matters when distributing and solving the system. Having linear factors is much simpler.