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Math Help - Long Division Problem - 2

  1. #1
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    Long Division Problem - 2

    \dfrac{x^{3} - 4x - 7}{x^{2}-x-6}

    First step?
    Last edited by Jason76; July 13th 2014 at 08:26 PM.
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  2. #2
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    Re: Long Division Problem - 2

    you can always just divide

    $x(x^2-x-6)=x^3-x^2-6x$

    $x^3-4x-7-(x^3-x^2-6x)=x^2+2x-7$

    $x^2+2x-7-(x^2-x-6)=3x-1$

    so your quotient is $(x+1)$ with remainder $3x-1$
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  3. #3
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    Re: Long Division Problem - 2

    Hello, Jason76!

    \dfrac{x^3 - 4x - 7}{x^2-x-6}

    . . \begin{array}{cccccccccc} &&&&&& x & + & 1 \\ && --&--&--&--&--&--& -- \\ x^2-x-6 & | & x^3 &&& - & 4x & - & 7 \\ && x^3 &-& x^2 &-& 6x \\ && --&--&--&--&-- \\ &&&& x^2 &+& 2x &-& 7 \\ &&&& x^2 &-& x &-& 6 \\ &&&& --&--&--&--&-- \\ &&&&&& 3x &-& 1 \end{array}


    Therefore: . \frac{x^3-4x-7}{x^2-x -6} \;=\;x+1 + \frac{3x-1}{x^2-x-6}
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