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Math Help - simplify radical

  1. #1
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    simplify radical

    8x^2-8x / 2x^3?
    What am I missing? I can't seem to figure out the bottom.
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  2. #2
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    Is it

    \frac{8x^2-8x}{2x^3}

    or

    8x^2-\frac{8x}{2x^3} ?

    For the first one u have:

    \frac{8x^2-8x}{2x^3}=\frac{8x(x-1)}{2x^3}=\frac{4(x-1)}{x^2}=\frac{4x-4}{x^2}=\frac{4}{x}-\frac{4}{x^2}

    For the second u will have

    8x^2-\frac{8x}{2x^3} =8x^2-\frac{4}{x^2}


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    Hush_Hush
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  3. #3
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    Quote Originally Posted by galanm View Post
    8x^2-8x / 2x^3?
    What am I missing? I can't seem to figure out the bottom.
    \dfrac{8x^2-8x}{2x^3} = \dfrac{2x(4x-4)}{2x\cdot x^2}=\dfrac{4(x-1)}{x^2}
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  4. #4
    A riddle wrapped in an enigma
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    Another look, using the greatest common factor first.

    \dfrac{8x^2-8x}{2x^3}=\frac{8x(x-1)}{2x^3}=\frac{\rlap{///}8^{\color{red}4}x(x-1)}{\rlap{//}2x^3_{\color{red}x^2}}=\frac{{\color{red}4}(x-1)}{{\color{red}x^2}}
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