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HELP! partial fractions driving me crazy

  1. Jan 5, 2007 #1
    Hi Guys, can any one help with this problem?

    resolve 3 -x
    (x^2 +3) (x + 3)

    The problem I have is with the x^2, when substituting numbers for x at the end to find A and B. I can only use -3
  2. jcsd
  3. Jan 5, 2007 #2


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    Why? Once you sub in -3 you will find A(or B, depending on your notation). Then sub in, say, 3. Since you know one variable, this will give you you an equation in only the variable you don't know!
  4. Jan 5, 2007 #3
    thanks cristo but I think I am missing something here.

    3 -x = A(x^2 +3) + B (x + 3)

    if x = -3 then A = 2

    do I now find the square root of 3 to make A (0)?

    I think I may have looked at this for too long now and confused myself!!!!!!
  5. Jan 5, 2007 #4


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    You can go here to have it all done out for you by a computer.
  6. Jan 5, 2007 #5


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    can be written as partial fractions as
    [tex]\frac{Ax+B}{x^2+3}+ \frac{C}{x+3}[/tex]
    You will need to solve for A, B, and C.
    Multiplying both sides by (x+3)(x2+3) you have
    (Ax+B)(x+3)+ C(x2+ 3)= 3-x

    Setting x= -3 give 9C= 6 or C= 2/3.

    Now let x be any two other numbers, say 0 and 1 for simplicity, to get two more equations for A and B.
  7. Jan 5, 2007 #6

    D H

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    Halls, setting x=-3 gives 12C=6 or C=1/2

    Equating like-power polynomial coefficients yields another way to attack these problems:

    x^2: A+C=0
    x^1: 3A+B=-1
    x^0: 3B+3C=3

    The result is three linear equations in three unknowns.
  8. Jan 5, 2007 #7

    Gib Z

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    Would it be incorrect of me to notice 3-x = -(x-3), cancel out the common factors to achieve -1/(x^2-3), and therefore ignoring when x=3? Makes it simpler.
  9. Jan 5, 2007 #8
    Yes it would be incorrect because there is no factor of (x-3) in the denominator. There is also no factor or (x2-3) in the denominator, I think you need to look at the original post again.
  10. Jan 5, 2007 #9

    Gib Z

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    Omg Thanks for correcting me on that, My god i must be going blind...
  11. Jan 8, 2007 #10
    Thankyou guys, you have all been very helpful, thanks for your time
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