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Homework Help: Abstract Algebra: Polynomials problem

  1. Jul 2, 2010 #1
    1. The problem statement, all variables and given/known data

    Let f(x)=x5-x2-1 [tex]\in[/tex] C and x1,...,x5 are the roots of f over C. Find the value of the symmetric function:

    (2x1-x14).(2x2-x24)...(2x5-x54)

    2. Relevant equations

    I think, that I have to use the Viete's formulas and Newton's Binomial Theorem.

    3. The attempt at a solution

    Unfortunately I can't solve the equation to the end. I will appreciate any help. Thank you :)
     
  2. jcsd
  3. Jul 2, 2010 #2
    hmm...all I can think of right now is that, since the constant of f(x) is -1, we know that the product of the roots is 1.

    So 2^5*x1x2x3x4x5 = 2^5 = 32, and (-x14)(-x24)(-x34)(-x44)(-x54) = -1.

    I know this doesn't solve your problem, but I hope it helps. I'll think about it some more in a bit.
     
    Last edited: Jul 2, 2010
  4. Jul 2, 2010 #3
    I'm pretty sure that there is a craftier way than I am about to introduce, but here goes.

    For any polynomial, [itex]f(x)[/itex], the product of the roots is the constant term of the polynomial. Also, the roots of [itex]f(x+a)[/itex] are related to the roots of [itex]f(x)[/itex] in the obvious way.

    Here, we have [itex]f(x) = x^5-x^2-1.[/itex]

    That's enough, but the factorization of [itex]x^4-2x[/itex] is crappy, so let's make it a little better.

    Note that if [itex]x_i^5-x_i^2-1=0,[/itex] then
    [tex]
    2x_i-x_i^4=\frac{x_i^2-1}{x_i}=\frac{(x_i-1)(x_i+1)}{x_i}
    [/tex]

    Then, to finish it off, your product is equivalent to
    [tex]
    \left(\prod_i (x_i-1)\right)\left(\prod_i (x_i+1)\right)/\left(\prod_i x_i\right) = (-3)(-1)/(-1)=-3
    [/tex]
    because the constant term of [itex]f(x)[/itex] is -1, [itex]f(x+1)[/itex] is -1, and [itex]f(x-1)[/itex] is -3.
     
  5. Jul 2, 2010 #4
    You may be right, but I still think your method was pretty neat. Nice job.
     
  6. Jul 2, 2010 #5

    Dick

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    Oooo. That is a really clever factorization. I was struggling with the 'crappy' factorization. Very nice. Except that the product of the roots is +1, like Raskolnikov said, isn't it? Not -1? Leaving a few details for oplik to clean up is a good idea.
     
  7. Jul 2, 2010 #6
    I double checked, I stand by my answer. The product of the roots is +1.

    But, in hindsight, this result from my original post needs explanation. It follows from one of Vi├Ęte's Laws. Namely, [tex] \prod_i x_i = (-1)^n \frac{a_0}{a_n} = (-1)^5 \frac{-1}{1} = 1. [/tex]
     
  8. Jul 3, 2010 #7
    Thank you, all of you. I think, I got it ;)
     
  9. Jul 3, 2010 #8
    Forgot the [itex](-1)^5[/itex]...Sorry.
     
  10. Jul 3, 2010 #9

    Dick

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    Don't apologize. You supplied the necessary hint. Correcting the details should be up to the OP. That's why I usually stop with the hint. Otherwise I make these detail mistakes.
     
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