Abstract Algebra: Polynomials problem

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Homework Statement



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

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

Homework Equations



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

The Attempt at a Solution



Unfortunately I can't solve the equation to the end. I will appreciate any help. Thank you :)
 
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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.
 
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I'm pretty sure that there is a craftier way than I am about to introduce, but here goes.

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

Here, we have f(x) = x^5-x^2-1.

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

Note that if x_i^5-x_i^2-1=0, then
<br /> 2x_i-x_i^4=\frac{x_i^2-1}{x_i}=\frac{(x_i-1)(x_i+1)}{x_i}<br />

Then, to finish it off, your product is equivalent to
<br /> \left(\prod_i (x_i-1)\right)\left(\prod_i (x_i+1)\right)/\left(\prod_i x_i\right) = (-3)(-1)/(-1)=-3<br />
because the constant term of f(x) is -1, f(x+1) is -1, and f(x-1) is -3.
 
tmccullough said:
I'm pretty sure that there is a craftier way than I am about to introduce, but here goes.

You may be right, but I still think your method was pretty neat. Nice job.
 
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.
 
Dick said:
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.

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 Viète's Laws. Namely, \prod_i x_i = (-1)^n \frac{a_0}{a_n} = (-1)^5 \frac{-1}{1} = 1.
 
Thank you, all of you. I think, I got it ;)
 
Forgot the (-1)^5...Sorry.
 
tmccullough said:
Forgot the (-1)^5...Sorry.

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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