Partial Fraction Decomposition (Irreducible Quadratic Denominators)
Key Questions

If you were trying to find the partial fraction decomposition of
#x/(x^29)# ,you would break up the denominator into
#(x+3)(x3)# , and the problem would be set up as follows:#x/((x+3)(x3))=A/(x+3)+B/(x3)# which would be simplified to be
#x=A(x3)+B(x3)# which is very easily solved.
However, if the problem were
#x/(x^481)# ,the denominator would factor into
#(x^2+9)(x^29)=(x^2+9)(x+3)(x3)# .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)(x3))=(Ax+B)/(x^2+9)+C/(x+3)+D/(x3)# 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. 
Imagine the partial fraction decomposition problem:
#(2x1)/(x^2x6)# Here, the denominator would simplify into
#(x3)(x+2)# so the decomposition would be set up as#(2x1)/((x3)(x+2))=A/(x3)+B/(x2)# However, when the denominator has a repeated factor, something slightly different happens.
Consider
#(x+3)/(x^2+4x+4)# Since
#x^2+4x+4=(x+2)^2# , you'd be tempted to set up the decomposition as#(x+3)/((x+2)(x+2))=A/(x+2)+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
#(x+3)/(x+2)^2=A/(x+2)+B/(x+2)^2# Which can indeed be solved.
An example with a third power:
#(3x2)/(2x1)^3=A/(2x1)+B/(2x1)^2+C/(2x1)^3# This is very similar for irreducible quadratic factors, except for that they take the form
#Ax+B# instead of just#A# . The same rules from earlier apply for "counting up" to the exponent.For example:
#(5x)/((x^2+4)^2(x3))=(Ax+B)/(x^2+4)+(Cx+D)/(x^2+4)^2+E/(x3)# 
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^24ac# and check if it is negative.#2x^2+3x+4# is irreducible because the discriminant is#932=23#
#3x^29x+5# is not irreducible because the discriminant is#8160=21#