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There are some polynomial with the above characteristic, and real roots of such polynomials cannot be found using rational number theorem and irrational conjugate theorem. The example of such function is this: $x^3-8x^2-2x+3=0$. If a integer coefficient polynomial has odd number of irrational roots, it implies that there is a irrational root without a distinct irrational conjugate.

Could you tell me how to solve a polynomial with a irrational root which doesn't have a distinct irrational conjugate using the example listed above? Also, please explain how come such an irrational root can exist and the condition in which such root appears.

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  • $\begingroup$ Does this link en.wikipedia.org/wiki/Irreducible_cubic give some help? (If I did not make mistakes the disriminant of you example polynomial is $D=7053$.) $\endgroup$ – gammatester Nov 7 '13 at 9:42
  • $\begingroup$ Thanks. That helped me a lot. Since this polynomial doesn't have a root which can be determined by Rational Root Test, it seems a Casus irreducibilis. I didn't know this concept, but your wikipedia's link taught me that. $\endgroup$ – Math.StackExchange Nov 8 '13 at 1:06

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