Polynomial Question

[jdom543]

Is it possible to have a polynomial that has a degree of 6 or higher (given the degree is even and isn't factor-able) with all imaginary roots or all irrational roots that is unsolvable?, not saying that it doesn't have any roots, but saying that it is impossible to find them.

Not sure if this is the right section for this but has to do with complex numbers so maybe. Nevertheless thank in advance for the help.

[stapel_eliz]

Creating a sixth-degree polynomial with all complex roots is one thing. Proving that there exist sixth-degree polynomials which are not solvable by radicals is quite another. (article)

What was the exact text of the question you need to answer? Thank you!

[jdom543]

Creating a sixth-degree polynomial with all complex roots is one thing. Proving that there exist sixth-degree polynomials which are not solvable by radicals is quite another. (article)

What was the exact text of the question you need to answer? Thank you!

Wasn't a question I was given, I was just wondering if it was impossible since we started talking in school about the rational root theorem and descarte's sign, etc.

Oddly enough my Stats teacher says you could solve them using interpolation and my Pre-Calc teacher says if one occurred you couldn't solve it....I was just curious.

[stapel_eliz]

It sounds like they're using two different meanings of "solve". "Interpolation" is a numerical method; it would be something similar to what your calculator or other software uses to find a decimal approximation of the zeroes. But it can be proven that there are polynomials which cannot be solved algebraically.

It would be like if there were no such thing as square roots. Would there then be a Quadratic Formula? So could you solve all quadratics algebraically, or would some have to be approximated from the graph?