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Math Help - Help with polynomial long division

  1. #1
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    Help with polynomial long division

    Here is the problem;
    (-30+6y^3-28y+y^2) divided by (2y-5)
    I have listed it in decending order as follows;
    6y^3+y^2 / -28-30
    I get 3y^2 and then I'm stuck...
    Any help or explanations would be much appreciated.
    I need to do this problem using long division and then do the self check.
    Last edited by mr fantastic; September 26th 2011 at 08:28 PM. Reason: Title.
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  2. #2
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    Re: I am lost, need help with polynomial long division

    Quote Originally Posted by sweetness4 View Post
    (-30+6y^3-28y+y^2) divided by (2y-5)
    I have listed it in decending order as follows;
    6y^3+y^2 / -28-30
    Your "decending order" makes no sense; should be:
    6y^3 + y^2 - 28y - 30

    Now you need to divide that by (2y - 5).
    We can't "teach" here; go here to learn:
    Polynomial Long Division
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  3. #3
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    Re: I am lost, need help with polynomial long division

    Wilmer thank you for the link, I will go look there. I did enter my decending order incorrectly (fraction bar shouldn't have been there). I've looked at about 4 sites so far and am still confused as to what to do with the y^2 when dividing. There have been a few other problems I was able to follow their steps and understand but this problem still has me confused. Thanks again for your help
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  4. #4
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    Re: I am lost, need help with polynomial long division

    Quote Originally Posted by sweetness4 View Post
    ...am still confused as to what to do with the y^2 when dividing.
    Not sure what you mean, like WHY the "y^2"
    Here's what happens at 1st step:
    Code:
             3y^2
            ------------------------
    2y - 5 / 6y^3 +   y^2 - 28y - 30
             6y^3 - 15y^2
             ------------
                    16y^2 - 28y
    See how the y^2 is "handled"? Hope that helps.
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  5. #5
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    Re: I am lost, need help with polynomial long division

    Yes, thats what I was asking. I'm understanding now. Thanks alot it was a big help
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