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Thread: hard problem

  1. February 18th 2009, 07:18 PM #1
    IronMan21
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    hard problem

    Let p(x)=x^3+ax^2+bx+c, where a, b, and c are complex numbers. Suppose that p(2009+9002{\pi}i)=p(2009)=p(9002)=0

    What is the number of nonreal zeros of x^12+ax^8+bx^4+c?


    That last part is x^12 btw
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  2. February 19th 2009, 01:40 AM #2
    red_dog
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    x^{12}+ax^8+bx^4+c=p(x^4)

    The real zeros of the equation are \pm\sqrt[4]{2009}, \ \pm\sqrt[4]{9002}

    So, there are 8 nonreal zeros.
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