Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I dont understand partial fractions for quadratic factors

  1. Jul 9, 2013 #1
    i understand the linear case...

    example..

    #/{(x+5)(x-4)} ----> A/(x+5) + B/(x-4)

    but i dont understand this..

    example..

    #/{(x^2+3)(x^2+9)}------>(Ax+B)/{(x-√3)(x+√3)} + (Cx+D)/(x^2+9)

    first of all... (x-√3)(x+√3)= x^2-3, which is nowhere in the original equation.. it's supposed to be x^2+3 no? so why then.. (x-√3)(x+√3)? and mainly.. i dont understand the whole.. "Ax+B" in the numerators for terms that have denominators on the bottom. why????

    why does it not make sense to do the following?...

    #/{(x^2+3)(x^2+9)}-----> A/(x^2+3) + B/(x^2+9)

    -------------------------------------------------------------

    why also.. for repeated linear factors do we do this?

    example..

    (2x+2)/{(x+5)^2}------> A/(x+5) + B/{(x+5)^2}

    .......................................................................................

    i know theres kind of a lot here to answer so i dont expect every person to answer everything so if you choose to answer specific cases that i asked about, please put the specific question that you are answering in quotes. thanks guys
     
  2. jcsd
  3. Jul 9, 2013 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    I don't know where you got your quadratic example from, but you have an incorrect factorization of the term x^2+3. As such, any partial fraction decomposition will necessarily be erroneous.

    The whole point of PF decomposition is to obtain factors whose numerators are of lesser degree than the denominators. This technique (PF) is used to facilitate the evaluation of integrals of rational functions.

    Once a partial fraction expression is obtained, it must be equivalent to the original expression. One checks this by adding the partial fractions together to see if the original expression results.
     
  4. Jul 9, 2013 #3

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

  5. Jul 9, 2013 #4

    symbolipoint

    User Avatar
    Homework Helper
    Education Advisor
    Gold Member

    The goal in Partial Fraction Decomposition is to find the individual fractions which were summed from which the given rational expression came from. There are Proper Rational Expressions and Improper Rational Expressions. When the degree of the numerator is equal or greater than the degree of the denominator, then this is an Improper Rational Expression, and you can perform division which gives one or more terms AND a remainder. This remainder is a Proper Fraction.

    When you study "ALGEBRA 1", you also learn to do sums or differences of rational expressions. You do the sum or difference, using if necessary the same "lowest" common denominator as you learned for ordinary base-ten numbers in fractions. You simplify, and you have your new rational expression. Partial Fraction Decomposition is just starting with the fraction, and reversing the process to find what the original summed fractions were.
     
  6. Jul 11, 2013 #5

    symbolipoint

    User Avatar
    Homework Helper
    Education Advisor
    Gold Member

    iScience said:


    You did not actually show the given portion being (x^2-9); but in fact showed the SUM of squares, so not factorable in Real numbers.

    You were hoping for some explanation about quadratic factors in partial fraction decomposition. You want to prepare to make numerators of the type, Ax+B so that you can make proper partial fractions from your given rational expression with quadratic denominators or quadratic factors in the denominator.

    You could have something like [itex]\frac{Ax+B}{Mx^2+Nx+P}[/itex] as one of your partial fractions, and it would be a PROPER fraction because the degree of the numerator is less than the degree of the denominator. You might, in fact, find either A=0 OR B=0 through the algebra steps, but you at least want to prepare for a possible binomial numerator. Note that a linear binomial numerator is degree 1 and the quadratic denominator is degree 2. This means the partial fraction would be a PROPER partial fraction.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: I dont understand partial fractions for quadratic factors
  1. Factoring a quadratic (Replies: 6)

  2. Partial Fractions (Replies: 7)

Loading...