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How do irreducible quadratic denominators complicate partial-fraction decomposition?

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

answered by: mason m
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