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Math Help - Polynomial equation

  1. #1
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    Polynomial equation

    If P(x) = 25x^6 - 25x^5 + 74x^4 - 24x^3 + 47x^2 + x - 2 is the complex number -i , a root of P(x)?

    it would really help me if you can post an explantion.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Polynomial equation

    Check to see if P(i)=0. To make it easier, use the fact that i^n=i^{n-4}.
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  3. #3
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    Re: Polynomial equation

    Quote Originally Posted by MarkFL2 View Post
    Check to see if P(i)=0. To make it easier, use the fact that i^n=i^{n-4}.
    how should I do that?
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Polynomial equation

    P(i)=25i^6-25i^5+74i^4-24i^3+47i^2+i-2

    i^6=i^2=-1

    i^5=i^1=i

    i^4=i^0=1

    i^3=i\cdot i^2=-i

    i^2=-1

    Now, using this, do you find P(i)=0?
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  5. #5
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    Re: Polynomial equation

    yes P(i) = 0
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  6. #6
    MHF Contributor MarkFL's Avatar
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    Re: Polynomial equation

    Yes, good work!
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  7. #7
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    Re: Polynomial equation

    so is that it? we proved that the complex number -i is a root of P(x)?
    Amazing Thank you so much.
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  8. #8
    MHF Contributor MarkFL's Avatar
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    Re: Polynomial equation

    Yes, that's all you have to do to see if a number is a root of a function. By the conjugate root theorem, we also know then that -i has to be a root as well, and so x^2+1 must be a factor of P(x).
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  9. #9
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    Re: Polynomial equation

    Thank you so much for helping MarkFL2! You are the most helpful. I also have 2 other examples I don't know how to do but they are about surjectivity/injectivity of a function, would you be able to help me?
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  10. #10
    MHF Contributor MarkFL's Avatar
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    Re: Polynomial equation

    I don't know much about that, I never studied it, but there are many helpful people here and I'm sure some of them have studied this.
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  11. #11
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    Re: Polynomial equation

    In general, if you want to find "rational" roots of an polynomial, try to do the following:

    For a continuous function, in the neighborhood of a root, both negative and positive values exist. Since polynomials satisfy these conditions, you can look for sign-changes and then do bi-section to narrow on roots. Typical problems have "nice" rational numbers, so, this is more like an exam tip. For example,

    (2x - 3)^3 = (4x^2 -12x + 9) * (2x - 3) = 8x^3 - 24x^2 + 18x - 12x^2 + 36x - 27 = 8x^3 - 36x^2 + 54x - 27.

    Consider its value at 0, which is -27, negative. Now consider it at 2, which is 1 that is positive. So, the zero must be somewhere in the middle. Now try 1, which is (0 + 2)/2, and so on... Remember, this can work if your roots are "nice" rational numbers, like 3/2, 5/4 etc... And, though it does not fail, you do not have infinite time to pursue this in exams, so, use it only to some extent, to get answer, otherwise, just switch to something else.

    In order to verify for higher x, pair successive terms, you end up with small and manageable values. Like, 8x^3 - 36x^2 + 54x - 27 = 4x^2(2x - 9) + 27(3x - 1).

    Salahuddin
    Maths online
    Last edited by Salahuddin559; October 26th 2012 at 06:48 AM.
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