TITLE: How to transform a general higher degree five or higher equation to normal form?
QUESTION [28 upvotes]: This is a continuation of How to solve fifth-degree equations by elliptic functions?
How to transform a general higher degree five or higher equation to normal form?  For example,  a quintic equation to Bring-Jerrard form?

REPLY [12 votes]: I believe Piezas' answer above covers pretty much every bit of the question, although it might not be entirely worthless to present a plausibility argument for such a transformation.
Start with the principle quintic $x^5 + ax^2 + bx + c$. It is evident that the first two elementary symmetric polynomials vanishes if the variables are all roots of the above quintic, i.e., 
$$x_1 + x_2 + x_3 + x_4 + x_5 = 0\tag{1}$$
$$x_1x_2 + x_1x_3 + x_1x_4 + x_1x_5 + x_2x_3 + x_2x_4 + x_2x_5 + x_3x_4 + x_3x_5 + x_4x_5 = 0 \tag{2}$$
For roots of the principle quintic $x_i$, $i = 1, \ldots, 5$
From these equalities, we see that the second-order newton polynomial of the roots vanishes too, i.e., 
$$x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 = 0\tag{3}$$
Let $f(x) = x^3 + px^2 - s_3/5$ and $g(x) = x^4 + qx^2 - s_4/5$ for $s_3$ and $s_4$ being the third and the fourth of the Newton-Girard symmetric polynomials. Applying $(1)$ and $(3)$, we have $\sum f = \sum g = 0$ where the argument is running through all the roots of the principle quintic.
Furthermore, 
$$\begin{align}\sum x f &= \sum_{i} x_i f(x_i) \\ &= \sum_{i} x_{i}^4 + p \sum_{i} x_i^3 - s_3\sum_{i} x_i \\ &= s_4 + ps_3 \end{align} \tag{4}$$
$$\begin{align}\sum x g &= \sum_{i} x_i g(x_i) \\ &= \sum_{i} x_{i}^5 + q \sum_{i} x_i^3 - s_4\sum_{i} x_i \\ &= s_5 + qs_3 \end{align} \tag{5}$$
Hence, setting $p = -s_4/s_3$ and $q = -s_5/s_3$ forces $\sum f = \sum g = 0$.
Now consider the shifted quintic derived by substituting $y = c_1 x + c_2 f + c_3 g$ into the principle quintic. We first calculate the first few Newton polynomials to get an idea of it's form :
$$ \sum_i y_i = c_1 \sum_i x_i + c_2 \sum_i f(x_i) + c_3 \sum_i g(x_i) = 0 \tag{6}$$
$$ \sum_i y_i^2 = c_1^2 \sum_i x_i^2 + c_2^2 \sum_i f(x_i)^2 + c_3^2 \sum_i g(x_i)^2 + 2c_1c_2 \sum_i x_i f(x_i) + 2c_2c_3 \sum_i f(x_i)g(x_i) + 2c_1c_3 \sum_i x_i g(x_i) = c_2^2 \sum_i f(x_i)^2 + c_3^2 \sum_i g(x_i)^2 + 2c_2c_3 \sum_i f(x_i)g(x_i) \tag{7}$$
Suppose we want to make $(7)$ vanish so that the coefficient of the term $x^3$ in transformed quintic becomes $0$. This implies we have to find a solution to the homogeneous quadratic equation in $(7)$ for some non-zero $c_2$ and $c_3$. It can be done by first dividing out through $c_3^2$ to transform it into a usual quadratic equation in $c_2/c_3$ and noting that 
$$\sum f^2 = -\frac{2s_4s_5}{s_3} + \frac{s_4^3}{s_3^2} - \frac{2s_3^2}{5} + \frac{2s_4s_2}{5} + \frac{s_3^2}{25} + s_6 \tag{8}$$
$$\sum fg = -\frac{s_4s_6}{s_3} - \frac{s_5^2}{s_3} - \frac{s_3s_4}{5} + \frac{s_4^2s_5}{s_3^2} - \frac{s_4s_3}{5} + \frac{s_4^2s_2}{5s_3} + \frac{s_5s_2}{5} + \frac{s_3s_4}{25} + s_7 \tag{9}$$
$$\sum g^2 = -\frac{2s_5s_6}{s_3} + \frac{s_5^2s_4}{s_3^2} - \frac{2s_4^2}{5} + \frac{2s_4s_5s_2}{5s_3} + \frac{s_4^2}{25} + s_8 \tag{10}$$
Where $s_1$ and similar notations are Newton-Girard polynomials, and can readily be expressed in terms of coefficients of the principal quintic. We ensure that the shifted quintic is the Bring-Jerrard form by noting that $\sum y^3 = 0$ is just a cubic equation in $c_1$ which can be satisfied for non-zero choice of $c_1$.
As a remark, I think it would have been more descriptive if I had given the coefficients in terms of Newton-Girard polynomials as well as a complete expression of $c_1, c_2$ and $c_3$ in terms the coefficients of the principal quintic but computing them in general form is very hard, even using some CAS at hand. This is one of the beauties of quintic. As you can see, even though a general quintic is not solvable by radicals, it takes a lot more radical than a quartic to even get reduced into a normal form. As for the Bring-Jerrard form, we see a solution to two quadratics (one from the principle form) and a solution to a cubic is required for this transformation, giving a total of three square roots and one cube root.