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Confused much, partial fractions

  1. Jul 29, 2008 #1
    .. oy, i'm just not sure how to find 3 constants!!
    Here is my problem:
    5x^2-4/(x-2)(x+2)(x-1) = A/(x-2)+B/(x+2)+C/(x-1)
    .. i got a bit of it done, but it's all wrong
    OH! and what am i supposed to do if the numerator of the first equation does not have any sort of variable with it??
    my next problem:
    3/r^2-4r-5 = A/r-5 + B/r+1
    Well, any help would be much appreciated ^^ :shy:
  2. jcsd
  3. Jul 29, 2008 #2


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    First, you can considerably simplify the problem by multiplying both sides of the equation by that common denominator:
    5x^2- 4= A(x+2)(x-1)+ B(x-2)(x-1)+ C(x-2)(x+2)

    Now you can do any of several things:

    You could multiply out the left side and equate "corresponding" coefficients.

    Or, since that is true for all x, you could just pick 3 values of x and get 3 equations to solve for A, B, and C.

    In particular, you will find taking x= -2, x= 1, and x= 2 are especially nice!

    For 3/r^2-4r-5 = A/r-5 + B/r+1, the same advice: multiply both sides by (r-5)(r+ 1) and you get 3= A(r+1)+ B(r-5). Now let r= -1 and r= 5.
  4. Jul 29, 2008 #3
    ? How did you pick those number? are they random, or no??
    Oh, and can i just say: THANK YOU!!
  5. Jul 29, 2008 #4


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    If you mean by choosing to test with x = -2, x = 1, and x=2, those numbers were clearly chosen so that all but one of the terms goes to zero. I must also add on that this is a remarkably good trick!

  6. Jul 29, 2008 #5
    ^^. thank you, i'll keep it in mind
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