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Math Help - algebraic fraction question HELP!!!

  1. #1
    Newbie Audriella's Avatar
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    algebraic fraction question HELP!!!

    ok so i deffinantly should know the answer to this question but my brain died about an hour ago and i do not see a resurrection in its future.

    i am begining to get very frusterated and am feeling stupider every time i try.

    so i will forever love you if someone could explain how to do this problem for me:

     (y^2+y+1)/(2y-2y^4)




    please and thank you
    Last edited by Audriella; January 5th 2008 at 03:48 PM. Reason: typo
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  2. #2
    Eater of Worlds
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    When faced with a problem like this, see if you can factor anything. That may lead to cancellations and simplification.

    \frac{y^{2}+y+1}{2y-2y^{4}}

    By the difference of two cubes, the denominator factors into 2y-2y^{4} = 2y(1-y^{3}) = 2y(1-y)(y^{2}+y+1)


    See now what to do?.
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  3. #3
    Newbie Audriella's Avatar
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    Quote Originally Posted by galactus View Post

    By the difference of two cubes, the denominator factors into 2y-2y^{4} = 2y(1-y^{3}) = 2y(1-y)(y^{2}+y+1)


    See now what to do?.

    Thank you so much for your help!

    but i dont understand how you go from [/tex]2y(1-y^{3})[/tex] to 2y(1-y)


    i am so sorry i really dont know where my brain is today.
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  4. #4
    Eater of Worlds
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    See what's in the parentheses?. That's the difference of two cubes

    x^{3}-y^{3}=(x-y)(x^{2}+xy+y^{2})

    (1)^{3}-(y)^{3}

    See it now?.
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  5. #5
    Newbie Audriella's Avatar
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    wow ok i am so sorry but i get it

    2y(1-y^{3}) = 2y[(1-y)(y^{2}+y+1)]
    so
    (y^{2}+y+1)/2y[(1-y)(y^{2}+y+1)] = 1/[2y(1-y)]


    right?
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  6. #6
    Eater of Worlds
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    Good work. You got it.
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