# Partial Fraction Decomposition (Irreducible Quadratic Denominators)

## Key Questions

• If you were trying to find the partial fraction decomposition of

$\frac{x}{{x}^{2} - 9}$,

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

$\frac{x}{\left(x + 3\right) \left(x - 3\right)} = \frac{A}{x + 3} + \frac{B}{x - 3}$

which would be simplified to be

$x = A \left(x - 3\right) + B \left(x - 3\right)$

which is very easily solved.

However, if the problem were

$\frac{x}{{x}^{4} - 81}$,

the denominator would factor into $\left({x}^{2} + 9\right) \left({x}^{2} - 9\right) = \left({x}^{2} + 9\right) \left(x + 3\right) \left(x - 3\right)$.

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:

$\frac{x}{\left({x}^{2} + 9\right) \left(x + 3\right) \left(x - 3\right)} = \frac{A x + B}{{x}^{2} + 9} + \frac{C}{x + 3} + \frac{D}{x - 3}$

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

• Imagine the partial fraction decomposition problem:

$\frac{2 x - 1}{{x}^{2} - x - 6}$

Here, the denominator would simplify into $\left(x - 3\right) \left(x + 2\right)$ so the decomposition would be set up as

$\frac{2 x - 1}{\left(x - 3\right) \left(x + 2\right)} = \frac{A}{x - 3} + \frac{B}{x - 2}$

However, when the denominator has a repeated factor, something slightly different happens.

Consider

$\frac{x + 3}{{x}^{2} + 4 x + 4}$

Since ${x}^{2} + 4 x + 4 = {\left(x + 2\right)}^{2}$, you'd be tempted to set up the decomposition as

$\frac{x + 3}{\left(x + 2\right) \left(x + 2\right)} = \frac{A}{x + 2} + \frac{B}{x + 2}$

However, this will not work. This is remedied by "counting up" to whatever the exponent is on the repeated factor. This would look like

$\frac{x + 3}{x + 2} ^ 2 = \frac{A}{x + 2} + \frac{B}{x + 2} ^ 2$

Which can indeed be solved.

An example with a third power:

$\frac{3 x - 2}{2 x - 1} ^ 3 = \frac{A}{2 x - 1} + \frac{B}{2 x - 1} ^ 2 + \frac{C}{2 x - 1} ^ 3$

This is very similar for irreducible quadratic factors, except for that they take the form $A x + B$ instead of just $A$. The same rules from earlier apply for "counting up" to the exponent.

For example:

$\frac{5 x}{{\left({x}^{2} + 4\right)}^{2} \left(x - 3\right)} = \frac{A x + B}{{x}^{2} + 4} + \frac{C x + D}{{x}^{2} + 4} ^ 2 + \frac{E}{x - 3}$

• Irreducible simply means that it can't be factored into real factors. So, an irreducible quadratic denominator means a quadratic that is in the denominator that can't be factored.

You can easily test a quadratic to check if it is irreducible. Simply compute the discriminant ${b}^{2} - 4 a c$ and check if it is negative.

$2 {x}^{2} + 3 x + 4$ is irreducible because the discriminant is $9 - 32 = - 23$
$3 {x}^{2} - 9 x + 5$ is not irreducible because the discriminant is $81 - 60 = 21$