Timeline for How to prove the following polynomial does not have root of a special form?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Dec 6, 2017 at 14:03 | history | edited | A. Mpi | CC BY-SA 3.0 |
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| Dec 6, 2017 at 13:40 | comment | added | Stanley Yao Xiao | In fact the polynomial $g(x) = x^4 - 7x^3 + 14x^2 - 8x + 1$ has Galois group isomorphic to $C_4$, and its roots can be found explicitly using the the identity $g(x) = x^2(x-2)^2 - 3x(x-1)(x-2) + (x-1)^2$. Since this polynomial is so special, I am very curious as to how it arose. What graph does this originate from, say? | |
| Dec 6, 2017 at 13:23 | answer | added | assaferan | timeline score: 5 | |
| Dec 6, 2017 at 13:09 | vote | accept | A. Mpi | ||
| Dec 6, 2017 at 13:03 | answer | added | Lior Bary-Soroker | timeline score: 4 | |
| Dec 6, 2017 at 13:00 | comment | added | Stanley Yao Xiao | This polynomial is solvable, because it is equal to $g(x^2)$ where $g(x) = x^4 - 7x^3 + 14x^2 - 8x + 1$; and quartic polynomials are always solvable. Thus you can find the roots explicitly and therefore determine whether they are or are not of the form that you ask. | |
| Dec 6, 2017 at 12:53 | comment | added | A. Mpi | Note that all roots of the above polynomial are real. | |
| Dec 6, 2017 at 12:43 | history | asked | A. Mpi | CC BY-SA 3.0 |