A peculiar factorization

Proving theorems in mathematics is frequently a tedious endeavour, because you’re usually just taking well-known techniques and applying them in a slightly different setting resulting in results that are ever more specialized. Often times you ‘know’ exactly what the answer should be and how the proof will go, expecting at most minor technical difficulties along the way.

However, once in a while your intuitions and expectations are shattered, or you encounter something that is mind-blowing. This is one such circumstance.

Recently I started working with the following family of polynomials

f_c(x) = x^6 - 3x^5 + cx^4 + (5 - 2c)x^3 + cx^2 - 3x + 1.$

Being a sextic, there is no general procedure to find the roots of this thing, and there is no “algebraic” reason why it should factor at all. I did not expect to be able to factor it at all, much less find an explicit procedure to solve the corresponding sextic equation using radicals. To understand these polynomials, I plotted them for several values of c and discovered the following facts, all unexpected, but not equally bizarre:

(1) The roots of f_c(x) are always all real or all complex, there is never any mixed roots. This is not surprising at all if you know where the family comes from, but without that knowledge, could be quite shocking.

(2) If the roots are all complex, then there is always a conjugate pair \alpha_1, \overline{\alpha_1} such that the real part of \alpha is 1/2. This, to me at least, was totally unexpected.

(3) If \alpha_2, \overline{\alpha_2} and \alpha_3, \overline{\alpha_3} are the two other pairs of complex roots, labelled so that \alpha_i have positive imaginary part, then in fact the imaginary parts of \alpha_2, \alpha_3 are equal. Moreover, the real parts of \alpha_2, \alpha_3 always sum to one! This is truly a wtf moment.

Using this data, one can in fact (after a lot of work and computer algebra) show that f_c(x) in fact always factors as follows:

f_c(x) = (x^2 - x + a)(x^2 - x/a + 1/a)(x^2 - (2 - 1/a)x + 1),

where a is a root of the cubic equation x^3 + (3 - c)x^2 + 3x - 1 = 0.

Since cubic equations are always solvable in radicals and quadratic equations are always solvable by radicals, it follows that f_c(x) is always solvable by radicals. Moreover, the procedure above is extremely explicit and simple to implement!

I have no explanation for why this happened. If you do, please let me know, I am very curious!


2 thoughts on “A peculiar factorization

  1. David Tweedle

    Hi Stanley.
    An important property of the polynomial you wrote down is that it is the same if you reverse the order of the coefficients.
    Algebraically: $x^6 f_c(1/x) = f_c(x)$
    So the roots come in pairs $ \alpha, 1/\alpha$ as well as the normal conjugate pairs you would expect.
    (Unless $f_c(1) = 0$ or $f_c(-1) = 0$ – maybe there would be a repeated root in that case)
    Then you can see that $\beta = \alpha + 1/\alpha$ is a degree $3$ extension of $\mathbb{Q}$, so $\beta$ can be written in terms of radicals, then $\alpha$ satisfies a quadratic equation over $\mathbb{Q}(\beta)$.
    I would be interested to know more.
    All the best – David

  2. Pingback: On binary forms | The Conscious Mathematician

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