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    the question asked me to show (2 / (x - 1) (x - 3)) in partial fractions which i did and got : (1 / (x - 3)) - (1 / (x - 1)) which is correct

    it then asked to use that answer to express (4 / ((x - 3)^2)((x - 1)^2)) in partial fractions,
    i thought i would just have to add a (1 / (x - 1)^2) and (1 / (x - 3)^2), which would give: (1 / (x - 3)) + (1 / (x - 3)^2) - (1 / (x - 1)) + (1 / (x - 1)^2),

    but the answer in the back is very slightly different:
    (1 / (x - 3)) - (1 / (x - 3)^2) + (1 / (x - 1)) + (1 / (x - 1)^2)

    can someone explain why i am wrong and explain a method?
    or is the book is wrong

    thanks in advance
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    anyone?
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    (Original post by goerigi)
    it then asked to use that answer to express (4 / ((x - 2)^2)((x - 1)^2)) in partial fractions,
    you sure that bit is correct?
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    (Original post by Freakonomics123)
    you sure that bit is correct?
    ohh that should be (x - 3)^2. typo! sorry
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    (Original post by Freakonomics123)
    you sure that bit is correct?
    so do you have any ideas
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    (4 / ((x - 3)^2)((x - 1)^2))

    equals the square of

    (2 / (x - 1) (x - 3)) = square of [(1 / (x - 3)) - (1 / (x - 1))]
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    (Original post by vc94)
    (4 / ((x - 3)^2)((x - 1)^2))

    equals the square of

    (2 / (x - 1) (x - 3)) = square of [(1 / (x - 3)) - (1 / (x - 1))]
    thanks but could you explain how (2 / (x - 1) (x - 3)) = square of [(1 / (x - 3)) - (1 / (x - 1))
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    (Original post by goerigi)
    thanks but could you explain how (2 / (x - 1) (x - 3)) = square of [(1 / (x - 3)) - (1 / (x - 1))
    the square of \frac{2}{(x-2)(x-3)} = the square of \frac{1}{(x-3)} - \frac{1}{(x-1)}

    because;

    \frac{2}{(x-2)(x-3)} = \frac{1}{(x-3)} - \frac{1}{(x-1)}
 
 
 
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