how to find coefficient c1, c2, c3, c4 of a polynomials of degree 4 from resolvent if not starting from standard resolvent of each degree
and use (y-x1...)(y-x2...)(y-x3...) and group theory
how to find corresponding c1, c2, c3, c4 of polynomial x^4+c4*x^3+c3*x^2+c2*x+c1 which c1, c2, c3, c4 are in terms of x1, x2, x3, x4  and  x^4+c4*x^3+c3*x^2+c2*x+c1  is separable which means (x-....)(x....)(x....)...?
update1:
x1,x2,x3,x4 may not be symmetric polynomials,
i see x1*x2 + x3*x4 as permutation group [[1,2],[3,4]]
or x1*x2*x3 + x4 as [[1,2,3],[4]]
david a. cox's book only have an example page 358 about working from x^4+c4*x^3+c3*x^2+c2*x+c1 to resolvent and then group, my question is reverse direction, which means from a set of groups and then resolvent and then one polynomial
if physicist start from group to work out a polynomial, how do they do in real practice?
update2
if not start from x1*x2+x3*x4  this permutation group  [[1,2],[3,4]] , and if start from for example, x1*x2*x3 + x4, resolvent is change to something like  (y-(x1*x2*x3 + x4)) * (y-.....etc, how to find back a polynomial for degree 4.
 A: I think we have established in the comments that part of the question is, given a function $f(x_1,x_2,\dots,x_n)$ of the roots of a polynomial, for example, $x_1x_2x_3+x_4$, how do we find the conjugates of this expression. 
The answer is, the conjugates are the expressions $f(x_{\sigma(1)},x_{\sigma(2)},\dots,x_{\sigma(n)})$, as $\sigma$ runs through all the elements of the group $S_n$. 
In the example, and writing elements of $S_n$ in cycle notation, $\sigma=(1\ 4)$ gives us $x_2x_3x_4+x_1$, $\sigma=(2\ 4)$ gives $x_1x_3x_4+x_2$, and $\sigma=(3\ 4)$ yields $x_1x_2x_4+x_3$ (and of course letting $\sigma$ be the identity gets us $x_1x_2x_3+x_4$). There are 20 other elements of $S_4$, but they just give us the same four conjugates over and over. 
The other part of the question is how to express $\prod_{\sigma}(y-f(x_{\sigma(1)},x_{\sigma(2)},\dots,x_{\sigma(n)}))$ in terms of the coefficients of the original polynomial. I will leave that to someone more proficient in the use of computer algebra systems than myself. 
