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Math Help - Expand Into Partial Fractions

  1. #1
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    Angry Expand Into Partial Fractions

    I know how to do these problems but this one is giving me some trouble:
    Expand into partial fractions

    5/(x+1)(x^2-1)

    ***Should I break it down into:
    1) [A/(x+1)]+[B/(x-1)]+[C/(x+1)] or
    2) [A/(x+1)]+[B/(x+1)^2]+[C/(x-1)]



    I tried it both ways and came up with different possibilities for the constraints:
    If Broken Down Like #1 from Above
    A+B+C=0
    2B = 0
    A+B-C=5

    If Broken Down Like #2 from Above
    A+C=0
    B+2C=0
    A+B-C=5
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  2. #2
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    Re: Expand Into Partial Fractions

    Which way gives you the LCD that is the denominator of the original expression?
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  3. #3
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    Re: Expand Into Partial Fractions

    I'm not sure because you can break down the original denominator (x+1)(x^2-1) into (x+1)(x+1)(x-1)
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    Re: Expand Into Partial Fractions

    Yes, and if you were to combine the 3 terms of both the choices you listed, which one would give you the correct common denominator?
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  5. #5
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    Re: Expand Into Partial Fractions

    Option 1 would give me that..right?
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  6. #6
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    Re: Expand Into Partial Fractions

    No it wouldn't. Since two of them have \displaystyle \begin{align*} x + 1 \end{align*} as the denominator, the LCD is \displaystyle \begin{align*} (x + 1)(x - 1) \end{align*}, NOT \displaystyle \begin{align*} (x + 1)^2(x - 1) \end{align*} as required.

    When you have repeated factors, you need one denominator to be \displaystyle \begin{align*} x + 1 \end{align*} and another to be \displaystyle \begin{align*} (x + 1)^2 \end{align*} in your partial fraction.
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    Re: Expand Into Partial Fractions

    so when I write it in the partial fractions I write it as:
    [A/(x+1)]+[B/(x+1)^2]+[C/(x-1)]

    ??
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  8. #8
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    Re: Expand Into Partial Fractions

    Yes
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  9. #9
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    Re: Expand Into Partial Fractions

    5/(x+1)(x^2-1) = A/(x+1) + B/(x+1)^2 + C/(x-1)
    multiply by LCD to get:

    5= A(x+1)(x-1) + B(x-1) + C(x+1)^2

    5= A(x^2-1) + B(x-1) + C(x^2+2x+1)

    5= Ax^2 - A + Bx - B + Cx^2 + 2Cx + C

    Group Common Terms:
    5= x^2(A+C) + x(B+2C) + (-A-B+C)

    Constraints:
    A+C=0
    B+2C=0
    A+B-C=5
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  10. #10
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    Re: Expand Into Partial Fractions

    Quote Originally Posted by rrooggeerr View Post
    5/(x+1)(x^2-1) = A/(x+1) + B/(x+1)^2 + C/(x-1)
    multiply by LCD to get:

    5= A(x+1)(x-1) + B(x-1) + C(x+1)^2

    5= A(x^2-1) + B(x-1) + C(x^2+2x+1)

    5= Ax^2 - A + Bx - B + Cx^2 + 2Cx + C

    Group Common Terms:
    5= x^2(A+C) + x(B+2C) + (-A-B+C)

    Constraints:
    A+C=0
    B+2C=0
    A+B-C=5
    This all looks fine. Now solve for A,B,C.
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  11. #11
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    Re: Expand Into Partial Fractions

    I get:
    A= -5/4
    B= -5/2
    C= 5/4

    I know something is wrong though because it doesn't make the third constraint true
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  12. #12
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    Re: Expand Into Partial Fractions

    never mind, i got it!!

    A = 5/4
    B= 5/2
    2= -5/4

    THANK YOU!
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  13. #13
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    Re: Expand Into Partial Fractions

    There is an alternate method for partial fraction decomposition you may be interested in:

    Heaviside cover-up method - Wikipedia, the free encyclopedia
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