Partial Fraction Decomposition (Irreducible Quadratic Denominators)

Key Questions

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

    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.

    mason m · 1 · Dec 20 2015
  • How do you find the partial fraction decomposition when you have repeated quadratic or linear factors?

    Imagine the partial fraction decomposition problem:

    #(2x-1)/(x^2-x-6)#

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

    #(2x-1)/((x-3)(x+2))=A/(x-3)+B/(x-2)#

    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:

    #(3x-2)/(2x-1)^3=A/(2x-1)+B/(2x-1)^2+C/(2x-1)^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(x-3))=(Ax+B)/(x^2+4)+(Cx+D)/(x^2+4)^2+E/(x-3)#

    mason m · 1 · Jan 25 2016
  • What is meant by an irreducible quadratic denominator?

    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^2-4ac# and check if it is negative.

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

    Amory W. · · Aug 21 2014

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