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I was teaching this morning Descartes' Rule of Signs to my Precalculus class and I wrote this polynomial on the board:

$f(x)=3x^5-2x^4+2x^3-3x^2+2x+1$

I found that there is 4 change of signs, therefore there is 4 positive real zeros, or 2 positive real zeros, or 0 positive real zero.

Then $f(-x)=3(-x)^5-2(-x)^4+2(-x)^3-3(-x)^2+2(-x)+1\\ -3x^5-2x^4-2x^3-3x^2-2x+1$ which means the there is only 1 change, therefore there is 1 negative real zero.

Now since irrational zeros and imaginary zeros come in pairs, the negative real zero must be rational. So I told them that and found the rational zeros as follows:

$p: \pm 1\\ q: \pm 1, \pm 3 \\ \frac{p}{q}: \pm 1, \pm \frac{1}{3}$,

so I used the synthetic division for both $-1,-\frac{1}{3}$ and found that none of the numbers give me zero. I was stunned in class and tried to use the WolframAlpha factor calculator and it gave me a different number as the factor that I just can't find.

So my question to you is, what is the value of the negative real zero and why?

Chappers
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Joshua Bonet
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  • Why should irrational roots come in pairs? See https://math.stackexchange.com/questions/1717553/cubic-polynomial-with-three-distinct-irrational-roots – egreg May 18 '18 at 21:35
  • I'm pretty sure the Galois group of this polynomial is going to be the full $S_5$, so the root is not expressible by radicals. – Chappers May 18 '18 at 21:50
  • "Now since irrational zeros ... come in pairs" Even if that were true, the pair could be positive. $x^2 -2$ has a positive and negative irrational root. – fleablood May 18 '18 at 22:32

1 Answers1

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We find the Galois group of the equation to examine its root structure. We can employ reduction modulo $p$ and Dedekind's theorem as suggested in this post: if a polynomial factors modulo $p$ into factors with degree $(h_1,h_2,h_3,\dotsc,h_k)$, there is an element of its Galois group with cycle type $h_1+h_2+h_3+\dotsb+h_k$ (see e.g. here for a reasonably simple proof). The discriminant factors into $3^3 \cdot 90067$, so we can find the cycle types by factorising modulo $p$ for various small $p \neq 3$. In particular,

  • $p=2$ gives $5$ (the factorisation being $1 + x^2 + x^5$),
  • $p=11$ gives $1+4$ (factorisation $3 (1 + x) (4 + 4 x + 6 x^2 + 2 x^3 + x^4)$),
  • $p=13$ gives $1+1+3$ (factorisation $3 (5 + x) (7 + x) (1 + 5 x + 9 x^2 + x^3)$).

This is enough: the corresponding group elements have order $5$, $4$ and $3$, so the order of the group must be divisible by their lowest common multiple, namely $60$. But this is larger than $5 \cdot 4 = 20$, which is the order of the largest solvable permutation group on $5$ elements. Hence whatever the Galois group is, it is not solvable.

(Indeed, because the discriminant is not a rational square, the Galois group is not contained in $A_5$. Since $A_5$ is the only proper subgroup of $S_5$ of order at least $60$, the Galois group must therefore be $S_5$. Another way to see this is to show that the polynomial factors to $3 (18 + x) (25 + x) (50 + x) (15 + 44 x + x^2)$ in $F_{59}$, so the Galois group contains a $5$-cycle and a transposition, which is sufficient to generate $S_5$.)

Since the Galois group is not solvable, the roots of the equation are not expressible by radicals. So there is no elementary way to write the negative real root, which has value approximately $-0.3110$.

Chappers
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  • with respect... is there another way to do this without abstract algebra or field theory? The OP teaches precalc – MaximusFastidiousIrreverence May 19 '18 at 00:07
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    Unfortunately I think the answer is no: showing that a polynomial is not solvable by radicals essentially requires the Galois group, or a low-tech version of it as in Abel's original impossibility proof for the general equation. There may be a more basic way to find the Galois group, by exhibiting some generating permutations, (or enough to rule out the solvable possibilities) but it's not a priori obvious how to produce those without plucking them out of the air. – Chappers May 19 '18 at 00:33
  • cool just wondering... – MaximusFastidiousIrreverence May 19 '18 at 00:35
  • I highly appreciate the answers to it, I have another question: What are the conditions for irrational roots to not come in pairs? – Joshua Bonet May 22 '18 at 15:20
  • @JoshuaBonet If the roots are complex, you have a map given by $i \mapsto -i$ that fixes them, which forces the non-real ones to come in conjugate pairs. But there is no such map for real irrationals. Indeed, it's quite easy to come up with cubics with three irrational roots, such as $8x^3-6x-1$ (which has the irrational numbers $\cos{(\pi/9)},\cos{(7\pi/9)}$ and $\cos{(13\pi/9)}$ as roots). – Chappers May 22 '18 at 21:20