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We can describe the roots of quadratic equations in terms of addition, subtraction, multiplication, division, and the square-root function $\sqrt a$ which computes a root of the special polynomial $x^2-a=0$. Similarly, the roots of cubic and quartic equations can be described with the aforementioned operations and the cube root $\root3\of a$.

In 1796, Bring found a method to express the roots of general quintic polynomials in terms of the aforementioned operations, the quintic root $\root5\of a$ and the bring radical $\mathop{\mathrm{Br}}(a)$ which computes a root of the special quintic polynomial $x^5+x+a$.

Can this scheme be extended? More specifically, can we express the roots of $n$-th degree polynomials in terms of roots of $r_q(a)$ of “special” polynomials of the form

$$r_q(a)=x_0\quad\hbox{such that }a+q_0+\sum_{k=1}^nq_kx^k=0?$$

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  • $\begingroup$ If you don't already know, it should be helpfully noted that the solutions to quintic equations are not always expressible in terms of radicals and arithmetic operations. $\endgroup$ – theREALyumdub Dec 4 '15 at 21:03
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    $\begingroup$ @theREALyumdub It isn't with just ordinary radicals $\root n\of m$, but with the Bring-radical, you always can (as far as I understood). $\endgroup$ – FUZxxl Dec 4 '15 at 21:04
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    $\begingroup$ Look up Galois theory. It was created for addressing questions like this one. I don't personally know the answer, but the question is essentially whether the combined splitting field of all polynomials of degree $n$ is a finite extension field of the rational numbers. $\endgroup$ – Paul Sinclair Dec 4 '15 at 23:36

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