roots of polynomials

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  1. chrisrobbo's Avatar
    • New Member
    • Posts: 11
    roots of polynomials
    Roots of polynomials

    a=alpha
    b=beta
    g=gamma (not sure how to get proper signs in)

    a^2 +b^2 +g^2 = (a+b+g)^2 -2(ab+bg+ag)

    My question what is the way to do it when (a^3 +b^3 +g^3) or each of them to power of four, or five etc...

    I have done a few papers and have seen this come up a few times.


    Thanks Chris.
  2. nuodai's Avatar
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    • TSR Legend
    Re: roots of polynomials
    You expand (\alpha + \beta + \gamma)^n and hope for the best.

    For instance

    (\alpha + \beta + \gamma)^3 = (\alpha^3 + \beta^3 + \gamma^3) + 3(\alpha^2 \beta + \alpha \beta^2 + \beta^2 \gamma + \beta \gamma^2 + \gamma^2 \alpha + \alpha \gamma^2) + 6\alpha \beta \gamma

    so rearranging gives

    \alpha^3 + \beta^3 + \gamma^3 = (\alpha + \beta + \gamma)^3 - 3(\alpha^2 \beta + \alpha \beta^2 + \beta^2 \gamma + \beta \gamma^2 + \gamma^2 \alpha + \alpha \gamma^2) - 6\alpha \beta \gamma

    and then you can simplify \alpha^2 \beta + \alpha \beta^2 + \beta^2 \gamma + \beta \gamma^2 + \gamma^2 \alpha + \gamma \alpha^2 to give

    \alpha^2( \beta + \gamma) + \beta^2(\alpha + \gamma) + \gamma^2(\alpha + \beta)

    then noticing for example that \beta + \gamma = (\alpha + \beta + \gamma) - \alpha and so on, we get that this is equal to

    (\alpha^2 + \beta^2 + \gamma^2)(\alpha + \beta + \gamma) - (\alpha^3 + \beta^3 + \gamma^3)

    so substituting this back in and rearranging, and then using your expression for \alpha^2+\beta^2+\gamma^2, will give you the answer.

    For \alpha^4+\beta^4+\gamma^4 it's a bit easier because you can write it as (\alpha^2)^2 + (\beta^2)^2 + (\gamma^2)^2 and then use your results for \alpha^2+\beta^2+\gamma^2 instead.

    I doubt you'll have to do all this stuff in an A-level exam (is this A-level?) but you might at degree-level. If it does come up at A-level then it will probably be made simpler either by a hint, given identity or a special case (e.g. where \alpha + \beta + \gamma = 0).

    Edit: I should mention, if you want to get good at doing things like the above, it's just a case of practice -- manipulate powers, take out factors, do lots of adding and subtracting, keeping in mind the goal of trying to write everything in terms of \alpha+\beta+\gamma and \alpha\beta + \beta\gamma + \gamma\alpha and \alpha \beta \gamma as possible. This is always possible (provided that you're given the right thing).
    Last edited by nuodai; 28-05-2012 at 13:51.
  3. h2shin's Avatar
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    • Posts: 610
    Re: roots of polynomials
    or there is a method where, if you have an equation with three different roots, the equation will be have a cubic order. And since they are roots they satisfy the equation you're given, so you substitute each root into the equation, then add all the equations together, then rearrange to leave the sum of cubes on one side and the rest on the other side and find the value.
  4. nuodai's Avatar
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    • TSR Legend
    Re: roots of polynomials
    (Original post by h2shin)
    or there is a method where, if you have an equation with three different roots, the equation will be have a cubic order. And since they are roots they satisfy the equation you're given, so you substitute each root into the equation, then add all the equations together, then rearrange to leave the sum of cubes on one side and the rest on the other side and find the value.
    That's certainly easier than the "blind" approach in my post! Nice work.

    (The blind approach is work keeping in mind though.)
  5. chrisrobbo's Avatar
    • New Member
    • Posts: 11
    Re: roots of polynomials
    Thank you very much for your answers, think i'll go with h2shin's answer little bit easier to do haha. Just tried an exam question and worked a treat.

    Also could you elaborate on what it means by cubic order?
  6. h2shin's Avatar
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    • Posts: 610
    Re: roots of polynomials
    (Original post by chrisrobbo)
    Thank you very much for your answers, think i'll go with h2shin's answer little bit easier to do haha. Just tried an exam question and worked a treat.

    Also could you elaborate on what it means by cubic order?
    just means the greatest power of x is 3. so x^2 - 9 is of a quadratic order, x^3 + x^2 - 9x is of a cubic order.
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