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

    xx2−9,

    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)=Ax+3+Bx−3

    which would be simplified to be

    x=A(x−3)+B(x−3)

    which is very easily solved.

    However, if the problem were

    xx4−81,

    the denominator would factor into (x2+9)(x2−9)=(x2+9)(x+3)(x−3).

    The x2+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(x2+9)(x+3)(x−3)=Ax+Bx2+9+Cx+3+Dx−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−1x2−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)=Ax−3+Bx−2

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

    Consider

    x+3x2+4x+4

    Since x2+4x+4=(x+2)2, you'd be tempted to set up the decomposition as

    x+3(x+2)(x+2)=Ax+2+Bx+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=Ax+2+B(x+2)2

    Which can indeed be solved.

    An example with a third power:

    3x−2(2x−1)3=A2x−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(x2+4)2(x−3)=Ax+Bx2+4+Cx+D(x2+4)2+Ex−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 b2−4ac and check if it is negative.

    2x2+3x+4 is irreducible because the discriminant is 9−32=−23
    3x2−9x+5 is not irreducible because the discriminant is 81−60=21

    Amory W. · · Aug 21 2014

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