Roots of a certain sixth order polynomial

Question

I am looking for the roots (or basically any information regarding them) of the sixth order polynomial $$p(x):=ax^6+(a+1)x^4+2bx^3-b^2$$ for positive, real constants $a,b$. Since $p(0)=-b^2<0$ and $\lim_{x\to\pm\infty}p(x)=+\infty$ we have at least two real roots $x_0,x_1$. Moreover, plotting the function for several parameters suggests, that the remaining four roots are non-real (i.e. two pairs of complex conjugate solutions).

Unfortunately this is as far as I got. I am especially interested in the complex solutions and their behavior for large parameters $a,b$. Any idea on how to tackle this problem is very welcome!

Answer 1

$\color{brown}{\textbf{Parametrization.}}$

Searching the given polynomial's representation in the form of $$P(a,b,x) =(a+2p)(x^6+t^6)+p(x^2+t^2)^3-3px^2(x^2+t^2)^2 + 2bx^3,\tag1$$ easily to get $$ax^6 -3pt^2 x^4 + 2b x^3 + (3p+a) t^6=ax^6+(a+1)x^4+2bx^3-b^2,$$

$$t^2=-\dfrac{a+1}{3p},\tag2$$ $$\quad (3p+a)(a+1)^3=27b^2p^3,\quad p^3=\dfrac{c^2}4(3p+a),$$ where $$c= \dfrac2b\sqrt{\dfrac{(a+1)^3}{27}}.\tag3$$

Then $$4\left(\dfrac pc\right)^3-3\,\dfrac pc = \dfrac ac.$$ Taking in account identities $$4s^3-3s = \cos(3\arccos s),$$ one can obtain $$p=c\cos\left(\dfrac23k\pi+\dfrac13\arccos\dfrac ac\right),\tag4$$ where $\;k\in\{-1,0,1\}\;$ provides the real positive result.

$\color{brown}{\textbf{Substitution.}}$

Assume $$y=x+\dfrac {t^2}x,\tag5$$ then $$x^3+\dfrac{t^6}{x^3}=\left(x+\dfrac {t^2}x\right)^3-3t^2\left(x+\dfrac {t^2}x\right)=y^3-3t^2y,$$ Therefore, the given polynomial $(1)$ presented in the form of $$P(a,b,x) = x^3Q\left(a,b,x-\dfrac{a+1}{3px}\right),\tag6$$ where $$Q(a,b,y) = (a+3p)y^3-3py^2+(a+2p)\dfrac{a+1}p y+2b.\tag7$$ Therefore, the given polynomial is reduced the cubic polynomial of the general form with the known constant cofficients.

$\color{brown}{\textbf{Example.}}$

Let $\;a=7,b=5,\;$ then $\;c\approx1.741\,859\,372\,646,\; p\approx2.172\,466\,644\,575.\;$

Then WA gives $\;y\approx-0.227\,483\,364\,391,\;$ with the right root $\;x_1\approx 1.000\,000\,000\,000.\;$

$\color{brown}{\textbf{Conclusions.}}$

From $(2)-(7)$ should the next:

• The roots of the given polynomial are solvable in the closed form via elementary functions;
• Identity $\;P\left(a,b,-\dfrac{a+1}{3px}\right)=-\dfrac{(a+1)^6}{(3px)^6}P\left(a,b,x\right)\;$ defines the correspondence between the values on the positive and negative arguments.

Answer 2

1. By Descartes rule of signs, the polynomial $$p(x):=ax^6+(a+1)x^4+2bx^3-b^2$$ has exactly one positive root. It is so because there is exactly one change of sign in $p(x).$
   Similarly, there is exactly one change of sign in $p(-x).$ Our polynomial has one negative root.
   Some bounds of real roots are found below.

2. Consider a monic polynomial $$q(x):=x^6+\frac{a+1}{a}x^4+2\frac{b}{a}x^3-\frac{b^2}{a}.\tag 1$$ Clearly, $q(x)$ and $p(x)$ share the roots.
   I assume that $a+1<2b$ and $b>2,$ from where $2\frac{b}{a}<\frac{b^2}{a}.$
   By Lagrange theorem 1.3., an upper bound for the positive real root is $\left(\frac{b^2}{a}\right)^{1/6}.$ Replacing $x$ by $-x$ in $(1)$ we get a lower bound for negative real root $-\left(2\frac{b}{a}\right)^{1/3}-\left(\frac{b^2}{a}\right)^{1/6}.$ Putting together, real roots $r_i$ satisfy $$-\left(2\frac{b}{a}\right)^{1/3}-\left(\frac{b^2}{a}\right)^{1/6}<r_i<\left(\frac{b^2}{a}\right)^{1/6}.$$

3. The remaining four roots are in conjugate pairs, $\;(\alpha\pm i\beta)\;$ and $\;(\gamma\pm i\delta).$ From the coefficients of $(1)$ we deduce that the sum of real parts of all six roots is $0,$ and the product of the roots is $-b^2\over a.$ The latest can be formulated as "the product of modules of the roots equals $b^2 \over a$".
   "Lagrange over $\mathbb{C}$" (Theorem 1.5 in the above mentionned article) gives an upper bound for the absolute values of the roots $w_i,$ real or complex: $$|w_i|<\left(2\frac{b}{a}\right)^{1/3}+\left(\frac{b^2}{a}\right)^{1/6}.$$

Answer 3

One thing to note is that there is no closed form solution in terms of radicals (as hinted in a comment).

Notice that if $p_{a,b}(x)$ admits an expression of solutions by radicals, then specialising $a$ and $b$ yields a radical expression for the roots over $\mathbb{Q}$. In particular, consider $$p_{1,2}(x) = x^6+ 2 x^4 + 4 x^3 - 4 \in \mathbb{Q}[x]$$

Recall from galois theory that there is a radical expression for the roots of $p_{1,2}(x)$, then the galois group of $p_{1,2}(x)$ must be solvable.

We may verify (by Eisenstein at $2$) that $p_{1,2}$ is irreducible, and furthermore checking the reduction mod $p$ for a few $p$ can eventually see that $\operatorname{Gal}(p_{1,2}(x)) = S_6$ - alternatively use computer algebra (e.g., GaloisGroup in Magma). Indeed $S_6$ is nonsolvable.

As a remark, if you are wondering about a radical expression over $\mathbb{C}$, a similar argument works while considering $p_{a,2}(x) \in \mathbb{C}(a)$, showing this galois group is not solvable.

Answer 4

$$ p(x) = ax^6+(a+1)x^4+2bx^3-b^2$$ I earlier thought the polynomial was solvable, until I checked the Galois group of $6x^6+7x^4+10x^3-25 = 0$ $$ G \in [720,-1,1,"S6"]$$ If the Galois group of any polynomial above quartic is the Symmetric group, then the polynomial is unsolvable, and you don't need to ask for its root, but instead their approximation

Your polynomial is in two variable, so lucky enough some would be solvable, I wrote a program on pari/gp to check

for( a=-20, 20, for(b= -20, 20 , f= ax^6+(a+1)x^4+2bx^3-b^2; if(a!=0&&polisirreducible(f)&&polgalois(f)!=[720,-1,1,"S6"], print(a, " ", b, " " , polgalois(f)))))

The result is just

$$ 9 , -1 , [72, -1, 1, "F_36(6):2 = [S(3)^2]2 = S(3) wr 2"]$$ $$9 , 1 , [72, -1, 1, "F_36(6):2 = [S(3)^2]2 = S(3) wr 2"]$$

So if $a= 9$, $b= 1$ , then $9x^6+10x^4+4x^3-1 = $ is solvable and irreducible among range $ -20 \le a,b \le 20$

Now in case you still want to see a formula to $ ax^6+(a+1)x^4+2bx^3-b^2$ I can get it for you but would require a resolvent of degree $15$

EDIT:

R < x > := PolynomialRing (RationalField ()); f := 9x^6+10x^4+2*x^3-1; G := GaloisGroup (f); G; IsSolvable (G);

Working on magma, the result is

Permutation group $G$ acting on a set of cardinality $6$ Order = $72$ = $2^3 \cdot 3^2$

$(1, 2)(3, 4)(5, 6)$

$(1, 3, 6)$

$ (1, 3)$

true