0

The task is to solve $$x^6+2x^5-5x^4+9x^3-5x^2+2x+1=0$$ by radicals. A hint is given to put $$u=t+1/t.$$ I tried factoring the polynomial and applying the rational root theorem, none gave me direction. I also don't know how to apply the hint.

Any suggestion or help will be very much appreciated. Thank you.

David K
  • 98,388
desperatemuch
  • 168
  • 3
    See http://math.stackexchange.com/questions/480102/quadratic-substitution-question-applying-substitution-p-x-frac1x-to-2x4x and http://math.stackexchange.com/questions/403025/equation-with-high-exponents – lab bhattacharjee Nov 27 '16 at 07:58

1 Answers1

2

Hint: the polynomial is self-reciprocal (or palindromic) and $x=0$ is obviously not a root, so divide by $x^3$ and regroup to get:

$$x^3 + \frac{1}{x^3} +2\left(x^2+\frac{1}{x^2}\right)-5\left(x+\frac{1}{x}\right)+9=0$$

Let $u = x + \frac{1}{x}$ so that $x^2 + \frac{1}{x^2} = u^2 - 2$ and $x^3+\frac{1}{x^3} = u^3 - 3u$ then the equation reduces to the cubic in $u$:

$$u^3+2 u^2 - 8 u + 5 = 0$$

which factors as:

$$(u-1)(u^2+3u-5)= 0$$

Each of the $3$ roots in $u$ gives a quadratic in $x$ to solve for the total of $6$ roots in $x$.

dxiv
  • 76,497
  • Thank you. This is the first time I heard about palindromic polynomials... – desperatemuch Nov 27 '16 at 08:17
  • 1
    So, this gives me the following roots: $$\frac{1+i\sqrt{3}}{2},$$ $$\frac{1-i\sqrt{3}}{2},$$ $$\frac{-3+\sqrt{29}+\sqrt{22-6\sqrt{29}}}{4},$$ $$\frac{-3+\sqrt{29}-\sqrt{22-6\sqrt{29}}}{4},$$ $$\frac{-3-\sqrt{29}+\sqrt{22+6\sqrt{29}}}{4},$$ $$\frac{-3-\sqrt{29}-\sqrt{22+6\sqrt{29}}}{4},$$ Hence, the given polynomial is solvable by radicals. – desperatemuch Nov 27 '16 at 08:22
  • @desperatemuch Those look right, except you should write $\sqrt{22-6\sqrt{29}}=,i,\sqrt{6\sqrt{29}-22}$ – dxiv Nov 27 '16 at 18:48
  • Why does it matter? I'm confused. – desperatemuch Nov 28 '16 at 22:48
  • @desperatemuch It's not critical in this particular case, but it's good practice for several reasons. (1) It's customary to make it obvious which roots are real vs. complex. (2) The square root of a positive number is understood as the plain radical virtually everywhere, while negative (or complex) values under the radical may introduce ambiguities depending on context, see for example this or that. – dxiv Nov 28 '16 at 23:06
  • @desperatemuch (3) For roots of higher order, the value may depend on the interpretation, see for example Why do I get an imaginary result for the cube root of a negative number?. – dxiv Nov 28 '16 at 23:07
  • @dixv So this implies that the splitting field for the given polynomial is $$\mathbb{Q}\left(i\sqrt{3},\sqrt{29}, i\sqrt{6\sqrt{29}-22},\sqrt{6\sqrt{29}+22}\right),$$ right? – desperatemuch Nov 30 '16 at 09:26
  • Don't know that it follows directly from the factorization. Guess you could check online with Magma or Sage, but that's not my favorite cup of tea. – dxiv Dec 01 '16 at 01:18