The polynomial

$p(x) = x^6-9x^4-4x^3+27x^2-36x-23$.

has at least one (real, irrational) root that is expressible by radicals (can you find it?).

Are all the roots of $p$ expressible by radicals and if so, how can one find the expressions?

  • $\begingroup$ My reasoning is that if there is one, there is a second. If there are two, then you can write the above as a product of a quadratic and a quartic ie all roots are expressible by radicals. $\endgroup$ – user88595 Jan 24 '14 at 0:58
  • $\begingroup$ Wolfram doesn't find that root : wolframalpha.com/input/… Not sure how we could! $\endgroup$ – user88595 Jan 24 '14 at 1:03
  • $\begingroup$ Typo perhaps? The equation $-x^6 - 9x^4-4x^3+27x^2-36x+23 = 0$ has for root 1. $\endgroup$ – user88595 Jan 24 '14 at 1:09
  • $\begingroup$ No typo there. I can give the root if you want. $\endgroup$ – ploosu2 Jan 24 '14 at 1:10
  • $\begingroup$ @user: I think your reasoning is suspect. $\endgroup$ – GEdgar Jan 24 '14 at 1:13

Maple says $$ 16(x^6-9x^4-4x^3+27x^2-36x-23) = \left( i\sqrt {3}\sqrt [3]{2}-2\,\sqrt {3}-\sqrt [3]{2}-2\,x \right) \\ \left( i\sqrt {3}\sqrt [3]{2}+2\,x+2\,\sqrt {3}+\sqrt [3]{2} \right) \left( i\sqrt {3}\sqrt [3]{2}-2\,\sqrt {3}+\sqrt [3]{2}+2\,x \right) \\ \left( i\sqrt {3}\sqrt [3]{2}+2\,\sqrt {3}-\sqrt [3]{2}-2\,x \right) \left( x+\sqrt {3}-\sqrt [3]{2} \right) \left( -x+\sqrt {3} +\sqrt [3]{2} \right) $$

  • $\begingroup$ Maple beats Wolfram $\endgroup$ – user88595 Jan 24 '14 at 1:57

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