Solving a 6th degree polynomial equation

Question

I have a polynomial equation that arose from a problem I was solving. The equation is as follows:

$$-x^6+x^5+2x^4-2x^3+x^2+2x-1=0 .$$

I need to find $x$, and specifically there should be a real value where $\sqrt3<x<\sqrt{2+\sqrt2}$, in accordance to the problem I am solving. I know that it would be possible for me to find approximations of the roots of the equation, but I would prefer to know the exact value of this specific root (i.e. with the answer as a surd, with nested surds if required). I am unable to do this as I do not know any method of solving polynomials of degree $> 4$.

If this cannot be done, could you tell me an approximate decimal value of $x$, or at least check that a solution exists within the range I have given (it is possible that I made an error earlier in my algebra).

Answer 1

The Rational Root Test shows that the only possible rational solutions are $\pm 1$. Substituting gives that $x = -1$ is one (but $x = 1$ is not), so polynomial long division gives $$p(x) = -(x + 1) q(x), \qquad q(x) := x^5 - 2 x^4 + 2 x^2 - 3 x + 1 .$$ for some quintic $q$. Computing that $q(-1) \neq 0$ is not a root of $q$, so if $q$ factors over $\Bbb Q$, it does so into an irreducible quadratic and an irreducible cubic, and since the constant term of $q$ is $1$, we need only find $3$ coefficients and a sign: $$q(x) = (x^2 + a x \pm 1) (x^3 + b x^2 + c x \pm 1).$$

Expanding and comparing like coefficients gives a simple quadratic system, yielding $$p(x) = -(x + 1)\underbrace{(x^2 - x + 1)(x^3 - x^2 - 2 x + 1)}_{q(x)} .$$ The discriminant of the quadratic is $-3 < 0$, so the real root you've identified must be a factor of the cubic. Since the cubic has no rational roots, one needs to use Cardano's Formula or the equivalent to extract it: The only root greater than $1$ is

$$x_0 := \frac{1}{6} \sqrt[3]{-28 + 84 i \sqrt 3} + \frac{14}3 \frac{1}{\sqrt[3]{-28 + 84 i \sqrt 3}} + \frac13 = 1.801937736\ldots .$$

Another method of solving the source problem as described by o.p. in the comments under Eric Towers' answer gives that the root is just $$x_0 = 2 \cos \frac\pi7 .$$ In retrospect, this form explains the bounds $\sqrt 3 < x_0 < \sqrt{2 + \sqrt{2}}$, i.e., $2 \cos \frac\pi6 < x_0 < 2 \cos \frac\pi8 ,$ in the question statement.

Answer 2

Using a complicated computer algebra system (Mathematica 10.4), we can get all the roots to this equation as radicals. Two roots are complex and the rest are all real (surprisingly).

Module[{roots},
  roots = Solve[-x^6 + x^5 + 2 x^4 - 2 x^3 + x^2 + 2 x - 1 == 0, x];
  Transpose[{
    N[x /. roots],
    FullSimplify[Element[x, Reals] /. roots],
    x /. roots
  }]
] // TableForm

• There are two complex roots at $\frac{1}{2} \pm \mathrm{i}\sqrt{3}$.
• There is a real root at $-1$.
• The other three are complicated and real: \begin{align} 1.80194\dots{} &= \frac{1}{3} \left(1+\frac{7^{2/3}}{\sqrt[3]{\frac{1}{2} \left(-1+3 \mathrm{i} \sqrt{3}\right)}}+\sqrt[3]{\frac{7}{2} \left(-1+3 \mathrm{i} \sqrt{3}\right)}\right), \\ -1.24698\dots{} &= \frac{1}{3}-\frac{7^{2/3} \left(1+\mathrm{i} \sqrt{3}\right)}{3\ 2^{2/3} \sqrt[3]{-1+3 \mathrm{i} \sqrt{3}}}-\frac{1}{6} \left(1-\mathrm{i} \sqrt{3}\right) \sqrt[3]{\frac{7}{2} \left(-1+3 \mathrm{i} \sqrt{3}\right)}, \\ 0.445042\dots{} &= \frac{1}{3}-\frac{7^{2/3} \left(1-\mathrm{i} \sqrt{3}\right)}{3\ 2^{2/3} \sqrt[3]{-1+3 \mathrm{i} \sqrt{3}}}-\frac{1}{6} \left(1+\mathrm{i} \sqrt{3}\right) \sqrt[3]{\frac{7}{2} \left(-1+3 \mathrm{i} \sqrt{3}\right)} \text{.} \end{align}

Conveniently, the first one is in the interval you require.

These three complicated roots are the roots of the same polynomial @Travis gets.

Looking at the Galois group structure, I believe we cannot dispense with complex numbers in these expressions.