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If $2 x^4 + x^3 - 11 x^2 + x + 2 = 0$, then find the value of $x + \frac{1}{x}$ ?

I would be very grateful if somebody show me how to factor this polynomial by hand, as of now I have used to Mathematica to get $(x-2) (2x - 1) (1 + 3x + x^2)$, but I am having trouble to factor this manually.

Best regards,

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Solving this type of quartic is easy, as it is symmetric. Also, see quasi-symmetric quartics (en.wikipedia.org/wiki/…). –  George Lowther Oct 27 '10 at 16:57
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The PlanetMath has more detail on symmetric quartics: planetmath.org/encyclopedia/SymmetricQuarticEquation.html –  George Lowther Oct 27 '10 at 17:07

2 Answers 2

up vote 8 down vote accepted

You don't need to factor the polynomial to find the value of $x + \frac{1}{x}$. Note that the given condition implies $\left( 2x^2 + \frac{2}{x^2} \right) + \left( x + \frac{1}{x} \right) - 11 = 0$ by dividing by $x^2$ and collecting symmetric terms. Now observe that $\left( x + \frac{1}{x} \right)^2 = x^2 + \frac{1}{x^2} + 2$...

But if you want to factor it by hand anyway, your best friend is the rational root test.

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Awesome,thank you very much :) –  Damir Frezivioski Oct 27 '10 at 15:40

HINT $\rm\displaystyle\quad \bigg(a\ \bigg(x+\frac{1}x\bigg)^2 + b\ \bigg(x+\frac{1}x\bigg)+c\bigg)\ x^2\: \ =\ \ a\ (x^4 + 1) + b\ (x^3+x) + (c+2a)\ x^2 $

The above allows you to determine the values of $\rm\:a,\:b,\:c\:$ by inspection. The same works for arbitrary even degree symmetric polynomials: $\ $ simply subtract $\rm\ a\ (x^2 +1)^k$ and recurse on what remains.

This is essentially a special case of Gauss's algorithm for rewriting symmetric polynomials in terms of elementary symmetric polynomials. It's one of the earliest known occurrences of Grobner basis rewriting, e.g. see p. 314 of Cox, Little, O'Shea:$\ $ Ideals, Varieties and Algorithms. In chapter 7 they present a generalization to compute the ring of invariants of a finite matrix group $\rm\: G \subset GL(n,k)\:$.

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