Homework Help: Abstract Algebra: Polynomials problem

1. Jul 2, 2010

oplik

1. The problem statement, all variables and given/known data

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)

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. Jul 2, 2010

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
3. Jul 2, 2010

tmccullough

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
$$2x_i-x_i^4=\frac{x_i^2-1}{x_i}=\frac{(x_i-1)(x_i+1)}{x_i}$$

Then, to finish it off, your product is equivalent to
$$\left(\prod_i (x_i-1)\right)\left(\prod_i (x_i+1)\right)/\left(\prod_i x_i\right) = (-3)(-1)/(-1)=-3$$
because the constant term of $f(x)$ is -1, $f(x+1)$ is -1, and $f(x-1)$ is -3.

4. Jul 2, 2010

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

5. Jul 2, 2010

Dick

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.

6. Jul 2, 2010

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, $$\prod_i x_i = (-1)^n \frac{a_0}{a_n} = (-1)^5 \frac{-1}{1} = 1.$$

7. Jul 3, 2010

oplik

Thank you, all of you. I think, I got it ;)

8. Jul 3, 2010

tmccullough

Forgot the $(-1)^5$...Sorry.

9. Jul 3, 2010

Dick

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.