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Math Help - another polynomial

  1. #1
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    another polynomial

    Thank you all for your help today! Here's another that's been stumping me:

    (2XY)(4XY)
    (12XY)
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  2. #2
    MHF Contributor red_dog's Avatar
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    \frac{(2x^{-1}y^2)(4x^2y^{-3})^{-2}}{(12x^{-2}y^{-2})^{-1}}=\frac{2x^{-1}y^2(2^2)^{-2}(x^2)^{-2}(y^{-3})^{-2}}{(2^2\cdot 3)^{-1}(x^{-2})^{-1}(y^{-2})^{-1}}=\frac{2x^{-1}y^22^{-4}x^{-4}y^6}{2^{-2}3^{-1}x^2y^2}=
    =\frac{2^{-3}x^{-5}y^8}{2^{-2}3^{-1}x^2y^2}=\frac{3}{2}x^{-7}y^6
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  3. #3
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    The answer in my booklet is

    3X
    2Y^7
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  4. #4
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    Sorry, I stand corrected:

    3X
    2Y^6
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  5. #5
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    Hello, tigersblood!

    The answer in my booklet is: . \frac{3x}{2y^7}

    Either you gave us the wrong problem . . . or your booklet is dreadfully wrong.

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  6. #6
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    I just checked and rechecked the problem. I copied it correctly, and the answer in the back is
    3x
    2y^6


    Aargh.
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