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Math Help - Another type of Partial Fractions

  1. #1
    Junior Member Cthul's Avatar
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    Another type of Partial Fractions

    The question starts as

    \frac{4x}{(1+x)(1-x)^{2}}

    I've gotten this far and don't know how to solve the rest.

    \frac{4x}{(1+x)(1-x)^{2}}\equiv\frac{A}{(1+x)}+\frac{B}{(1-x)}+\frac{C}{(1-x)^{2}}

    4x=A(1-x)^{2}+B(1-x)(1+x)+c(1+x)

    Substitute
    x=1

    2=c

    Then I don't know how to solve the rest, when I substitute an x, I get a zero for both, it's completely different from this.
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  2. #2
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    Quote Originally Posted by Cthul View Post
    The question starts as

    \frac{4x}{(1+x)(1-x)^{2}}

    I've gotten this far and don't know how to solve the rest.

    \frac{4x}{(1+x)(1-x)^{2}}\equiv\frac{A}{(1+x)}+\frac{B}{(1-x)}+\frac{C}{(1-x)^{2}}

    4x=A(1-x)^{2}+B(1-x)(1+x)+c(1+x)

    Substitute
    x=1

    2=c


    Then I don't know how to solve the rest, when I substitute an x, I get a zero for both, it's completely different from this.
    Hi Cthul,

    now that you have found C, place x=-1 to bypass B and discover A.

    Then as you have both A and C, you can place x=0 to find B.
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  3. #3
    Junior Member Cthul's Avatar
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    It won't work, it doesn't when I sub in zero, I get 2 variables.
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  4. #4
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    Quote Originally Posted by Cthul View Post
    It won't work, it doesn't when I sub in zero, I get 2 variables.
    Try substituting x = -1 first to find A.

    Once you have A, then you can let x = 0 to find B.



    My preferred method is expanding all the brackets, equating coefficients of like powers of x and solving the system.
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  5. #5
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    Quote Originally Posted by Cthul View Post
    It won't work, it doesn't when I sub in zero, I get 2 variables.
    No,

    substitute in the values you obtained for A and C before finding B.

    x=1, C=2.

    x=-1, 4A=-4, A=-1

    x=0, 0=A+B+C=-1+B+2=B+1, so B=-1
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  6. #6
    Junior Member Cthul's Avatar
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    I thought when x is substituted with zero, you'd get... (1-(-1))^{2} which makes .... oh, I see. Thanks. The signs get confusing.
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  7. #7
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    Quote Originally Posted by Cthul View Post
    I thought when x is substituted with zero, you'd get... (1-(-1))^{2} which makes .... oh, I see. Thanks. The signs get confusing.
    Thinking of subtracting negative numbers as the difference between temperatures can be helpful...

    5-2 is the difference between 5 and 2 degrees, which is 3 degrees.

    There is a 6 degree difference between between 5 and -1,
    5 degrees down to zero and another degree down to -1.
    That's a drop of 6 degrees.

    So 5-(-1) is that difference which we know is 6,
    subtracting the lower temperature from the bigger one.

    Since 5+1=6, we can say 5--1=5+1,
    so 1--1=1+1 and so on...

    -2-3 is subtract 2, then subtract 3, which is subtract 5 or -(2+3)

    -2--3=-2-(-3), the difference between -2 degrees and -3 degrees which is 1 degree
    as -2 is greater than -3,

    so -2--3=1=3-2 or -2+3 etc
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  8. #8
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    Partial fractions appear to be a very interesting problem.
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