1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.