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Math Help - quartic polynomial roots

  1. #1
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    quartic polynomial roots

    let  \alpha , \beta,  \gamma,  \delta be the roots of the polynomial equation.


     x^{4}+5x^{3}+3x^{2}-x+2=0

    calculate  \alpha^{2} +\beta^{2} +\gamma^{2} + \delta^{2}

    ..my thoughts... polynomial can be written as  (x-\alpha)(x-\beta)(x-\gamma)(x-\delta) expand this and equate? but will this be general for every polynomial irrespective of the signs of the coefficients? thanks appreciate its alot of latex!
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  2. #2
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    Quote Originally Posted by skystar View Post
    let  \alpha , \beta,  \gamma,  \delta be the roots of the polynomial equation.


     x^{4}+5x^{3}+3x^{2}-x+2=0

    calculate  \alpha^{2} +\beta^{2} +\gamma^{2} + \delta^{2}
    Note,
    (a+b+c+d)^2 = (a^2+b^2+c^2+d^2)+2(ab+ac+ad+bc+bd+cd).

    Now by Viete's theorem, a+b+c+d = - 5 and ab+ac+ad+bc+bd+cd = 3.

    Now solve for a^2+b^2+c^2+d^2.
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  3. #3
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    still confused i cant seem to put it in context with my example. thanks
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  4. #4
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    Quote Originally Posted by skystar View Post
    still confused i cant seem to put it in context with my example. thanks
    Let us see. Say the polynomial factors as (x-a)(x-b)(x-c)(x-d).
    Now expand the polynomial,
    x^4-(a+b+c+d)x^3
    +(ab+ac+ad+bc+bd+cd)x^2
    -(abc+abd+acd+bcd)x +abcd

    Now line up the coeffcients,
    a+b+c+d = -5
    ab+ac+ad+bc+bd+cd = 3
    abc+abd+acd+bcd = 1
    abcd=2

    EDIT: Mistake fixed.
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  5. #5
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    small typo

    <br />
    " alt="-(abc+abd+acd+bcd)x^3 +abcd
    " />

     -(abc+abd+acd+bcd)x +abcd
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