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Math Help - Algebraic division

  1. #1
    Newbie
    Joined
    Sep 2007
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    Algebraic division

    I have this function:

    f(x) = x + \frac{3}{x - 1} - \frac{12}{x^2 + 2x - 3}

    We've been asked to show that the above equation can be written like this:

    \frac{x^2 + 3x + 3}{x + 3}

    I just can't seem to get it and i can't figure out why!

    Could someone show me how it's done please?

    Many thanks.
    Last edited by psychocostin; September 20th 2007 at 09:54 AM.
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  2. #2
    Senior Member DivideBy0's Avatar
    Joined
    Mar 2007
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    Melbourne, Australia
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    x+\frac{3}{x-1}-\frac{12}{x^2+2x-3}

    =x+\frac{3}{x-1}-\frac{12}{(x-1)(x+3)}

    =\frac{x(x-1)(x+3)+3(x+3)-12}{(x-1)(x+3)}

    =\frac{x^3+2x^2-3x+3x+9-12}{(x-1)(x+3)}

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

    For x^3+2x^2-3, test 1:
    (1)^3+2(1)^2-3=0

    Hence, x-1 is a factor of x^3+2x^2-3. Using polynomial long division, you get the other factor to be x^2+3x+3.

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

    =\frac{x^2+3x+3}{x+3}
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  3. #3
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    Thanks a lot! I've been looking at that for days and got as far as where you found x -1 to be a factor.
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