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?
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$\begingroup$ 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. $\endgroup$– user88595Jan 24, 2014 at 0:58
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$\begingroup$ Wolfram doesn't find that root : wolframalpha.com/input/… Not sure how we could! $\endgroup$– user88595Jan 24, 2014 at 1:03
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$\begingroup$ Typo perhaps? The equation $-x^6 - 9x^4-4x^3+27x^2-36x+23 = 0$ has for root 1. $\endgroup$– user88595Jan 24, 2014 at 1:09
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$\begingroup$ No typo there. I can give the root if you want. $\endgroup$– ploosu2Jan 24, 2014 at 1:10
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1$\begingroup$ @user: I think your reasoning is suspect. $\endgroup$– GEdgarJan 24, 2014 at 1:13
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1 Answer
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Maple says $$ 16(x^6-9x^4-4x^3+27x^2-36x-23) = \left( i\sqrt {3}\sqrt [3]{2}-2\,\sqrt {3}-\sqrt [3]{2}-2\,x \right) \\ \left( i\sqrt {3}\sqrt [3]{2}+2\,x+2\,\sqrt {3}+\sqrt [3]{2} \right) \left( i\sqrt {3}\sqrt [3]{2}-2\,\sqrt {3}+\sqrt [3]{2}+2\,x \right) \\ \left( i\sqrt {3}\sqrt [3]{2}+2\,\sqrt {3}-\sqrt [3]{2}-2\,x \right) \left( x+\sqrt {3}-\sqrt [3]{2} \right) \left( -x+\sqrt {3} +\sqrt [3]{2} \right) $$
