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Partial Fractions

  1. May 23, 2004 #1
    Right, I'm gettin irritated by these :confused: , hehe, I need some expert quidance on how to do all kindsa questions with these, mainly the more complicated 1's where u can't just sub in values of x to get 0. Lotsa input will be appreciated :smile:
  2. jcsd
  3. May 23, 2004 #2


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    Some examples, specifics will help
  4. Jun 1, 2004 #3


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    "partial fractions" is generally a calculus subject. Was there a reason for posting this under "linear and abstract algebra"?
  5. Jun 1, 2004 #4
    Partial fractions themselves arent really even calculus at all!! Integrating perhaps...
  6. Jun 4, 2004 #5
    Its not really a calculus technique, its simply a technique that has its major application in the simplification of rational integrals. However the technique of breaking up a rational function into partial fractions has no calculus involved whatsoever, its all algebra.

    And yes specific examples of the problems giving you difficulty would be helpful to make us helpful.
  7. Jun 4, 2004 #6
    I love partial fractions. Bring on some examples. :)
  8. Jun 5, 2004 #7


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    Should I be worried? :frown:
  9. Jun 5, 2004 #8

    Tom Mattson

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    Do we really need specific examples? There are only two kinds of factors that you can encounter: linear and irreducible quadratic. They may be varied, and they may be repeated, but there is a simple procedure for handling them.

    Linear factors of the form (ax+b)m are treated as follows:


    Irreducible quadratic factors of the form (ax2+bx+c)n are treated as follows:


    Once you put all your terms in, you equate it to the original rational function and multiply both sides by the least common denominator. After that, the "brute force" way to do it would be to expand everything, collect like terms, and equate the coefficients of each power of x.

    edit: fixed a superscript bracket
  10. Aug 3, 2004 #9
    The above poster explained it well, remember the following too:

    -Decide which form the partial fraction will take. Write out the constants/form next to the original sum and form a common denominator that matches the original fraction. You can then solve.
    -Always factorise the bottom of the fraction first if at all possible.
    -If the degree of the denominator is greater than the degree (highest coefficient of x) of the numerator this is an improper fraction and first you need to divide by the numerator, ie polynomial long division. Any remainder stays over the original denominator, the value you get from dividing can be written cleanly and seperately, without any denominator. The remainder may then be simplified into further partial fractions if applicable.

    A few examples:

    Linear factors in the denominator:
    (4x^3 + 2x^2 + 3x + 5) / (x+1) (x-2) would take the form A/(x+1), B(x-2)
    Thus, A(x-2) + B(x+1) would = (4x^3 + 2x^2 + 3x + 5)
    You can first let x = -1 to find A, then let x = 2 to find B.

    Quadratic factors in the denominator:
    (x^2 - 5x + 1)/ (x^2 +1)(x-2) would take the form, (Ax+b)/(x^2 +1) + C(x-2)
    Hence, (Ax+b)(x-2) + C(x^2 +1) = (x^2 - 5x + 1)
    You could first let x =2, to find the value of C.
    You would then substitute in the value of C and multiply out before equating the coefficients. A simultaneous equation may occur but this is easily solved.

    Repeated factors in the denominator:
    2(x^2 - 2x - 1)/ (x+1) (x-1)^2 would take the form A/(x+1) + B/(x-1) + C/(x-1)^2
    You can then solve as normal.

    Hope this helps.
  11. Aug 3, 2004 #10
    Also, you simply substitute in different values of x to make the result in brackets equal to 0. This normally cancels one or two of the constants and can greatly facilitate the process. As the partial fraction format is an identity it is true for any value of x.
    For example, A(x-2) + B(x+1) = x, you would first cancel the A constant by making x=2, this would enable you to find B. You could then enter x=-1, in a similar manner.
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