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# How do irreducible quadratic denominators complicate partial-fraction decomposition?

How do irreducible quadratic denominators complicate partial-fraction decomposition?

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.

answered by: mason m
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