Descartes' Rule of Signs, Factor and Rational Zeros Theorem

Question

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?

Answer 1

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 $-0.3110$.