# Principal form of a polynomial

In mathematics and, more specifically, in theory of equations, the principal form of an irreducible polynomial of degree at least three is a polynomial of the same degree n without terms of degrees n−1 and n−2, such that each root of either polynomial is a rational function of a root of the other polynomial.
The principal form of a polynomial can be found by applying a suitable Tschirnhaus transformation to the given polynomial.

## Definition

Let
    $f(x)=x^{n}+a_{1}x^{n-1}+\cdots +a_{n-1}x+a_{n}$
be an irreducible polynomial of degree at least three.
Its principal form is a polynomial
    $g(y)=y^{n}+b_{3}y^{n-3}+\cdots +b_{n-1}y+b_{n},$
together with a Tschirnhaus transformation of degree two
    $\varphi (x)=x^{2}+\alpha x+\beta$
such that, if r is a root of f, $\phi (r)$ is a root of ⁠$g$⁠.
Expressing that ⁠$g$⁠ does not has terms in ⁠$y^{n-1}$⁠ and ⁠$y^{n-2}$⁠ leads to a system of two equations in ⁠$\alpha$⁠ and ⁠$\beta$⁠, one of degree one and one of degree two. In general, this system has two solutions, giving two principal forms involving a square root. One passes from one principal form to the secong by changing the sign of the square root.

## Cubic case
### Tschirnhaus transformation with three clues

The Tschirnhaus transformation always transforms one polynome into another polynome of the same degree but with a different unknown variable. The mathematical relation of the new variable to the old variable shall be called the Tschirnhaus key. This key is a polynome that has to satisfy special criteria about its coefficients. To fulfill these criteria a separate equation system of several unknowns has to be solved. The singular equations of that system are important clues that are composed in tables that are formulated in the following sections:
This is the given cubic equation:
    $x^{3}-ax^{2}+bx-c=0$
Following quadratic equation system shall be solved:
    ${\text{First clue}}$ $au+3v+a^{2}-2b=0$ $\mathrm {Second\,clue}$ $bu^{2}-3v^{2}+abu-3cu-2ac+b^{2}=0$ $\mathrm {Third\,clue}$ $w=cu^{3}+acu^{2}+bcu+v^{3}+c^{2}$
So exactly this Tschirnhaus transformation appears:
    $(x^{2}+ux+v)^{3}-w=0$
The solutions of this system, accurately the expression of u, v and w in terms of a, b and c can be found out by the substitution method. It means for instance, the first of the three chested equations can be resolved after the unknown v and this resolved equation can be inserted into the second chested equation, so that a quadratic equation after the unknown u appears. In this way, from the three to be solved unknowns only one unknown remains and can be solved directly. By finding out the first unknown, the further unknowns can be found out by inserting the computed unknown. By detecting all these unknown coefficients the mentioned Tschirnhaus key and the new polynome resulting from the mentioned transformation can be constructed. In this way the Tschirnhaus transformation is done.

### Cubic calculation examples

The quadratic radical components of the coefficients are identical to the square root terms appearing along with the Cardano theorem and therefore the Cubic Tschirnhaus transformation even can be used to derive the general Cardano formula itself.
Plastic constant:
    $x^{3}-x-1=0$ Tschirnhaus transformation in the mentioned pattern: ${\biglx^{2}+({\tfrac {1}{6}}{\sqrt {69}}-{\tfrac {3}{2}})x-{\tfrac {2}{3}}{\bigr ]}^{3}-{\tfrac {23}{54}}(3{\sqrt {69}}-23)=0$ Doing the cube root: ${\color {green}{\color {blueviolet}x^{2}}+({\tfrac {1}{6}}{\sqrt {69}}-{\tfrac {3}{2}})x-{\tfrac {2}{3}}={\tfrac {1}{6}}{\sqrt {23}}{\bigl12{\sqrt {3}}-4{\sqrt {23}}\,{\bigr ]}^{1/3}}$ Squaring of the previous equation: ${\biglx^{2}+({\tfrac {1}{6}}{\sqrt {69}}-{\tfrac {3}{2}})x-{\tfrac {2}{3}}{\bigr ]}^{2}-(x+{\tfrac {1}{3}}{\sqrt {69}}-3)(x^{3}-x-1)={\tfrac {23}{36}}{\bigl12{\sqrt {3}}-4{\sqrt {23}}\,{\bigr ]}^{2/3}$ Simplifying the left side of the balance: ${\color {blue}{\color {blueviolet}-({\tfrac {1}{2}}{\sqrt {69}}-{\tfrac {23}{6}})x^{2}}+{\tfrac {1}{9}}{\sqrt {69}}\,x+{\tfrac {1}{3}}{\sqrt {69}}-{\tfrac {23}{9}}={\tfrac {23}{36}}{\bigl12{\sqrt {3}}-4{\sqrt {23}}\,{\bigr ]}^{2/3}}$ Eliminating the x^2 term by combining the colored equations: ${\color {Cyan}x={\tfrac {1}{6}}{\sqrt {3}}{\bigl{\sqrt[{3}]{12{\sqrt {3}}+4{\sqrt {23}}}}+{\sqrt[{3}]{12{\sqrt {3}}-4{\sqrt {23}}}}\,{\bigr )}}$
Supergolden constant:
    $x^{3}-x^{2}-1=0$ Tschirnhaus transformation in the mentioned pattern: ${\biglx^{2}+({\tfrac {1}{2}}{\sqrt {93}}-{\tfrac {11}{2}})x-{\tfrac {1}{6}}{\sqrt {93}}+{\tfrac {3}{2}}{\bigr ]}^{3}-{\tfrac {31}{18}}(29{\sqrt {93}}-279)=0$ Doing the cube root: ${\color {green}{\color {blueviolet}x^{2}}+({\tfrac {1}{2}}{\sqrt {93}}-{\tfrac {11}{2}})x-{\tfrac {1}{6}}{\sqrt {93}}+{\tfrac {3}{2}}}={\tfrac {1}{6}}{\sqrt {93}}{\bigl116-12{\sqrt {93}}\,{\bigr ]}^{1/3}$ Squaring of the previous equation: ${\biglx^{2}+({\tfrac {1}{2}}{\sqrt {93}}-{\tfrac {11}{2}})x-{\tfrac {1}{6}}{\sqrt {93}}+{\tfrac {3}{2}}{\bigr ]}^{2}-(x+{\sqrt {93}}-10)(x^{3}-x^{2}-1)={\tfrac {31}{12}}{\bigl116-12{\sqrt {93}}\,{\bigr ]}^{2/3}$ Simplifying the left side of the balance: ${\color {blue}{\color {blueviolet}-({\tfrac {29}{6}}{\sqrt {93}}-{\tfrac {93}{2}})x^{2}}+({\tfrac {10}{3}}{\sqrt {93}}-31)x-({\tfrac {31}{6}}-{\tfrac {1}{2}}{\sqrt {93}})={\tfrac {23}{36}}{\bigl12{\sqrt {3}}-4{\sqrt {23}}\,{\bigr ]}^{2/3}}$ Eliminating the x^2 term by combining the colored equations: ${\color {Cyan}x={\tfrac {1}{6}}{\sqrt[{3}]{116+12{\sqrt {93}}}}+{\tfrac {1}{6}}{\sqrt[{3}]{116-12{\sqrt {93}}}}+{\tfrac {1}{3}}}$
Tribonacci constant:
    $x^{3}-x^{2}-x-1=0$ Tschirnhaus transformation in the mentioned pattern: ${\biglx^{2}+({\tfrac {1}{4}}{\sqrt {33}}-{\tfrac {9}{4}})x-{\tfrac {1}{12}}{\sqrt {33}}-{\tfrac {1}{4}}{\bigr ]}^{3}-{\tfrac {11}{72}}(19{\sqrt {33}}-99)=0$ Doing the cube root: ${\color {green}{\color {blueviolet}x^{2}}+({\tfrac {1}{4}}{\sqrt {33}}-{\tfrac {9}{4}})x-{\tfrac {1}{12}}{\sqrt {33}}-{\tfrac {1}{4}}}={\tfrac {1}{6}}{\sqrt {33}}{\bigl19-3{\sqrt {33}}\,{\bigr ]}^{1/3}$ Squaring of the previous equation: ${\biglx^{2}+({\tfrac {1}{4}}{\sqrt {33}}-{\tfrac {9}{4}})x-{\tfrac {1}{12}}{\sqrt {33}}-{\tfrac {1}{4}}{\bigr ]}^{2}-(x+{\tfrac {1}{2}}{\sqrt {33}}-{\tfrac {7}{2}})(x^{3}-x^{2}-x-1)={\tfrac {11}{12}}{\bigl19-3{\sqrt {33}}\,{\bigr ]}^{2/3}$ Simplifying the left side of the balance: ${\color {blue}{\color {blueviolet}-({\tfrac {19}{24}}{\sqrt {33}}-{\tfrac {33}{8}})x^{2}}+({\tfrac {3}{4}}{\sqrt {33}}-{\tfrac {11}{4}})x+{\tfrac {13}{24}}{\sqrt {33}}-{\tfrac {77}{24}}={\tfrac {11}{12}}{\bigl19-3{\sqrt {33}}\,{\bigr ]}^{2/3}}$ Eliminating the x^2 term by combining the colored equations: ${\color {Cyan}x={\tfrac {1}{3}}{\sqrt[{3}]{19+3{\sqrt {33}}}}+{\tfrac {1}{3}}{\sqrt[{3}]{19-3{\sqrt {33}}}}+{\tfrac {1}{3}}}$

### Cardano formula

The direct solving of the mentioned system of three clues leads to the Cardano formula for the mentioned case:
    $x^{3}-ax^{2}+bx-c=0$
    $x={\tfrac {1}{3}}a+{\tfrac {1}{3}}{\bigla^{3}-{\tfrac {9}{2}}ab+{\tfrac {27}{2}}c-{\sqrt {{\bigla^{3}-{\tfrac {9}{2}}ab+{\tfrac {27}{2}}c{\bigr )}^{2}-{\bigla^{2}-3b{\bigr )}^{3}}}\,{\bigr ]}^{1/3}$
    $+{\tfrac {1}{3}}{\bigla^{3}-{\tfrac {9}{2}}ab+{\tfrac {27}{2}}c+{\sqrt {{\bigla^{3}-{\tfrac {9}{2}}ab+{\tfrac {27}{2}}c{\bigr )}^{2}-{\bigla^{2}-3b{\bigr )}^{3}}}\,{\bigr ]}^{1/3}$

## Quartic case
### Tschirnhaus transformation with four clues

This is the given quartic equation:
    $x^{4}-ax^{3}+bx^{2}-cx+d=0$
Now this quadratic equation system shall be solved:
    $\mathrm {First\,clue}$ $at+4u+a^{2}-2b=0$ $\mathrm {Second\,clue}$ $bt^{2}-6u^{2}+abt-3ct-2ac+b^{2}+2d=0$ $\mathrm {Third\,clue}$ $v=ct^{3}+act^{2}-4dt^{2}-3adt+bct+4u^{3}-2bd+c^{2}$ $\mathrm {Fourth\,clue}$ $w=dt^{4}-u^{4}+uv+adt^{3}+bdt^{2}+cdt+d^{2}$
And so accurately that Tschirnhaus transformation appears:
    $(x^{2}+tx+u)^{4}-v(x^{2}+tx+u)+w=0$

### Quartic calculation examples

The Tschirnhaus transformation of the equation for the Tetranacci constant contains only rational coefficients:
    $x^{4}-x^{3}-x^{2}-x-1=0$
    $y=x^{2}-3x$
    $y^{4}-11y-41=0$
In this way following expression can be made about the Tetranacci constant:
    $x^{2}-3x=({\tfrac {41}{3}})^{1/4}{\sqrt {\sinh {\bigl{\tfrac {1}{3}}\operatorname {arsinh} ({\tfrac {363}{26896}}{\sqrt {123}}){\bigr ]}}}-$
    $-({\tfrac {41}{3}})^{1/4}{\bigl \{\tfrac {11}{4}}({\tfrac {3}{41}})^{3/4}{\sqrt {\operatorname {csch} {\bigl{\tfrac {1}{3}}\operatorname {arsinh} ({\tfrac {363}{26896}}{\sqrt {123}}){\bigr ]}}}-\sinh {\bigl{\tfrac {1}{3}}\operatorname {arsinh} ({\tfrac {363}{26896}}{\sqrt {123}}){\bigr ]}{\bigr \}}^{1/2}$
That calculation example however does contain the element of the square root in the Tschirnhaus transformation:
    $x^{4}+x^{3}+x^{2}-x-1=0$
    $y=x^{2}+{\tfrac {1}{5}}(19+4{\sqrt {21}})x+{\tfrac {1}{5}}(6+{\sqrt {21}})$
    $y^{4}-{\tfrac {1}{125}}(38267+8272{\sqrt {21}})y-{\tfrac {1}{625}}(101277{\sqrt {21}}+463072)=0$

### Special form of the quartic

In the following we solve a special equation pattern that is easily solvable by using elliptic functions:
    $x^{4}-6x^{2}-8{\sqrt {S^{2}+1}}\,x-3=0$
    $Q=q{\bigl \\tanh {\bigl{\tfrac {1}{2}}\operatorname {arsinh} (S){\bigr ]}{\bigr \}}=q{\biglS\div ({\sqrt {S^{2}+1}}+1){\bigr ]}$
    $x={\frac {3\,\vartheta _{01}(Q^{3})^{2}}{\vartheta _{01}(Q)^{2}}}$
These are important additional informations about the elliptic nome and the mentioned Jacobi theta function:
    $q(\varepsilon )=\exp {\bigl-\pi K({\sqrt {1-\varepsilon ^{2}}})\div K(\varepsilon ){\bigr ]}$
    $\vartheta _{01}(r)=\sum _{k=-\infty }^{\infty }(-1)^{k}r^{k^{2}}=\prod _{n=1}^{\infty }(1-r^{2n})(1-r^{2n-1})^{2}$
Computation rule for the mentioned theta quotient:
    ${\frac {3\,\vartheta _{01}\{q[\kappa ^{3}\div ({\sqrt {\kappa ^{6}+1}}+1)]^{3}\}^{2}}{\vartheta _{01}\{q[\kappa ^{3}\div ({\sqrt {\kappa ^{6}+1}}+1)]\}^{2}}}={\sqrt {2{\sqrt {\kappa ^{4}-\kappa ^{2}+1}}-\kappa ^{2}+2}}+{\sqrt {\kappa ^{2}+1}}$
Accurately the Jacobi theta function is used for solving that equation.
Now we create a Tschirnhaus transformation on that:
    $x^{4}-6x^{2}-8{\sqrt {S^{2}+1}}\,x-3=0$
    $y=x^{2}-2({\sqrt {S^{2}+1}}-S)x-3$
    $y^{4}+64\,S^{2}(4S^{2}+1-4S{\sqrt {S^{2}+1}})y-384\,S^{3}({\sqrt {S^{2}+1}}-S)=0$

### Elliptic solving of principal quartics

Given principal quartic equation:
    $x^{4}+\psi x-\omega =0$
If this equation pattern is given, the modulus tangent duplication value S can be determined in this way:
    $\psi ^{4}{\bigl384\,S^{3}({\sqrt {S^{2}+1}}-S){\bigr ]}^{3}=\omega ^{3}{\bigl64\,S^{2}(4S^{2}+1-4S{\sqrt {S^{2}+1}}){\bigr ]}^{4}$
The solution of the now mentioned formula always is in pure biquadratic radical relation to psi and omega and therefore it is a useful tool to solve principal quartic equations.
    $Q=\exp {\bigl \langle }-\pi K{\bigl \\operatorname {sech} {\bigl{\tfrac {1}{2}}\operatorname {arsinh} (S){\bigr ]}{\bigr \}}\div K{\bigl \\tanh {\bigl{\tfrac {1}{2}}\operatorname {arsinh} (S){\bigr ]}{\bigr \}}{\bigr \rangle }=$
    $=q{\bigl \\tanh {\bigl{\tfrac {1}{2}}\operatorname {arsinh} (S){\bigr ]}{\bigr \}}=q{\bigl \\tanh {\bigl{\tfrac {1}{2}}\operatorname {artanh} (S\div {\sqrt {S^{2}+1}}){\bigr ]}{\bigr \}}$
And this can be solved in that way:
    $x={\frac {\omega [64\,S^{2}(4S^{2}+1-4S{\sqrt {S^{2}+1}})]}{\psi [384\,S^{3}({\sqrt {S^{2}+1}}-S)]}}{\biggl{\frac {9\,\vartheta _{01}(Q^{3})^{4}}{\vartheta _{01}(Q)^{4}}}-2({\sqrt {S^{2}+1}}-S){\frac {3\,\vartheta _{01}(Q^{3})^{2}}{\vartheta _{01}(Q)^{2}}}-3{\biggr ]}$

### Calculation examples with elliptic solutions

Now this solving pattern shall be used for solving some principal quartic equations:
First calculation example:
    $x^{4}+x-1=0$
    $Q=q{\bigl \\tanh {\bigl{\tfrac {1}{2}}\operatorname {artanh} ({\tfrac {31}{100}}+{\tfrac {1}{300}}{\sqrt {849}}-{\tfrac {1}{300}}{\sqrt {386{\sqrt {849}}-1902}}){\bigr ]}{\bigr \}}$
    $x={\frac {4}{{\sqrt {2{\sqrt {849}}+18}}-6}}{\biggl{\frac {9\,\vartheta _{01}(Q^{3})^{4}}{\vartheta _{01}(Q)^{4}}}-{\frac {1}{4}}{\sqrt {32+2{\sqrt {6{\sqrt {849}}-54}}}}\,{\frac {3\,\vartheta _{01}(Q^{3})^{2}}{\vartheta _{01}(Q)^{2}}}-3{\biggr ]}$
Second calculation example:
    $x^{4}+2x-1=0$
    $Q=q{\bigl \\tanh {\bigl{\tfrac {1}{2}}\operatorname {artanh} ({\tfrac {1}{10}}+{\tfrac {1}{30}}{\sqrt {129}}-{\tfrac {1}{30}}{\sqrt {26{\sqrt {129}}-102}}){\bigr ]}{\bigr \}}$
    $x={\frac {2}{{\sqrt {2{\sqrt {129}}+18}}-6}}{\biggl{\frac {9\,\vartheta _{01}(Q^{3})^{4}}{\vartheta _{01}(Q)^{4}}}-{\frac {1}{2}}{\sqrt {8+2{\sqrt {6{\sqrt {129}}-54}}}}\,{\frac {3\,\vartheta _{01}(Q^{3})^{2}}{\vartheta _{01}(Q)^{2}}}-3{\biggr ]}$
Third calculation example:
    $x^{4}+5x-3=0$
    $Q=q{\bigl \\tanh {\bigl{\tfrac {1}{2}}\operatorname {artanh} ({\tfrac {239}{5092}}+{\tfrac {75}{5092}}{\sqrt {881}}-{\tfrac {5}{5092}}{\sqrt {11618{\sqrt {881}}-112750}}){\bigr ]}{\bigr \}}$
    $x={\frac {4}{{\sqrt {2{\sqrt {881}}+50}}-10}}{\biggl{\frac {9\,\vartheta _{01}(Q^{3})^{4}}{\vartheta _{01}(Q)^{4}}}-{\frac {1}{4}}{\sqrt {32+10{\sqrt {2{\sqrt {881}}-50}}}}\,{\frac {3\,\vartheta _{01}(Q^{3})^{2}}{\vartheta _{01}(Q)^{2}}}-3{\biggr ]}$

## Quintic case
### Synthesis advice for the quadratic Tschirnhaus key

This is the given quintic equation:
    $x^{5}-ax^{4}+bx^{3}-cx^{2}+dx-e=0$
That quadratic equation system leads to the coefficients of the quadratic Tschirnhaus key:
    $\mathrm {First\,clue}$ $as+5t+a^{2}-2b=0$ $\mathrm {Second\,clue}$ $bs^{2}-10t^{2}+abs-3cs-2ac+b^{2}+2d=0$
By polynomial division that Tschirnhaus transformation can be made:
    $(x^{2}+sx+t)^{5}-u(x^{2}+sx+t)^{2}+v(x^{2}+sx+t)-w=0$

### Calculation examples

This is the first example:
    $x^{5}-x^{4}-x^{2}-1=0$
    $y=x^{2}-{\tfrac {1}{4}}(19-{\sqrt {265}})x-{\tfrac {1}{20}}({\sqrt {265}}-15)$
    $y^{5}+{\tfrac {1}{80}}(24455-1501{\sqrt {265}})y^{2}-{\tfrac {1}{160}}(5789{\sqrt {265}}-93879)y-{\tfrac {1}{4000}}(5393003{\sqrt {265}}-87785025)=0$
And this is the second example:
    $x^{5}+x^{4}+x^{3}+x^{2}-1=0$
    $y=x^{2}+{\tfrac {1}{3}}({\sqrt {30}}-3)x+{\tfrac {1}{15}}{\sqrt {30}}$
    $y^{5}-{\tfrac {1}{45}}(465-61{\sqrt {30}})y^{2}+{\tfrac {2}{45}}(1616-289{\sqrt {30}})y-{\tfrac {1}{1125}}(33758{\sqrt {30}}-183825)=0$

### Solving the principal quintic via Adamchik and Jeffrey transformation

The mathematicians Victor Adamchik and David Jeffrey found out how to solve every principal quintic equation. In their essay Polynomial Transformations of Tschirnhaus, Bring and Jerrard they wrote this way down. These two mathematicians solved this principal form by transforming it into the Bring Jerrard form. Their method contains the construction of a quartic Tschirnhaus transformation key. Also in this case that key is a polynome in relation to the unknown variable of the given principal equation y that results in the unknown variable z of the transformed Bring Jerrard final equation.
For the construction of the mentioned Tschirnhaus transformation key they executed a disjunction of the linear term key coefficient in order to get a system that solves all other terms in a quadratic radical way and to only solve a further cubic equation to get the coefficient of the linear term key coefficient.
In their essay they constructed the quartic Tschirnhaus key in this way:
    Steps of solving the principal Given principal equation: $y^{5}-uy^{2}+vy-w=0$ Tschirnhaus key: $z=y^{4}+{\color {crimson}\alpha }y^{3}+{\color {green}\beta }y^{2}+{\color {orange}\gamma }y+{\color {blue}\delta }$ Bring Jerrard final form: $z^{5}+\lambda z-\mu =0$
In order to do the transformation the mathematicians Adamchik and Jeffrey constructed a special equation system that generates the coefficients of the cubic, quadratic and absolute term coefficients of the Tschirnhaus key. Along with their essay of polynomial transformations, these coefficients can be found out by combining the expressions of the quartic and cubic term of the final Bring Jerrard form that are equal to zero because in this way the Bring Jerrard equation form is defined.
By combining these expressions of the zero valued quartic and cubic term of the Bring Jerrard final form, an equation system for the unknown Tschirnhaus key coefficients can be constructed. And this resulting equation system can be simplified by combining the equation clues in the essay into each other. In this way the following simplified equation system of two unknown key coefficients can be set up:
    Equation system of two unknows $4v{\color {crimson}\alpha }-3u{\color {green}\beta }=5w$ $3u^{2}{\color {crimson}\alpha }^{2}-3uv{\color {crimson}\alpha }+25w{\color {crimson}\alpha }{\color {green}\beta }-10v{\color {green}\beta }^{2}=5uw-2v^{2}$
On the basis of the essay by Adamchik and Jeffrey, the just mentioned equation system of two unknowns results from setting the zero valued quartic coefficient of the Bring Jerrard final form into the zero valued cubic coefficient and eliminating all terms of the linear key coefficients and absolute key coefficients. In other words, eliminating all gamma and delta terms. In this way you get the red colored cubic term coefficient and the green colored quadratic term coefficient of the Tschirnhaus key. The mentioned zero valued quartic coefficient of the Bring Jerrard final form is accurately this one here:
    $3u{\color {crimson}\alpha }+5{\color {blue}\delta }-4v=0$
Solving the zero valued quartic coefficient of the Bring Jerrard final form leads directly to the blue colored absolute term coefficient of the Tschirnhaus key.
    ${\color {blue}\delta }={\frac {4}{5}}v-{\frac {3}{5}}u{\color {crimson}\alpha }$
And for receiving the orange colored linear term coefficient of the Tschirnhaus key, the zero valued quadratic coefficient of the Bring Jerrard final form must be solved after the mentioned linear term coefficient of the Tschirnhaus key. And accurate that is done by solving this cubic equation:
    $u{\color {orange}\gamma }^{3}+(5w{\color {crimson}\alpha }-4v{\color {green}\beta }+3u^{2}){\color {orange}\gamma }^{2}+(uv{\color {crimson}\alpha }^{2}+6uw{\color {crimson}\alpha }+5w{\color {green}\beta }^{2}-8uv{\color {green}\beta }+3u^{3}+vw){\color {orange}\gamma }+$
    $+u^{3}{\color {crimson}\alpha }^{3}+vw{\color {crimson}\alpha }^{3}-2u^{2}{\color {green}\beta }^{3}+2uv{\color {crimson}\alpha }{\color {green}\beta }^{2}-2u^{2}{\color {crimson}\alpha }^{2}{\color {blue}\delta }+10{\color {blue}\delta }^{3}-$
    $-4u^{2}v{\color {crimson}\alpha }^{2}+vw{\color {crimson}\alpha }{\color {green}\beta }+3uv{\color {crimson}\alpha }{\color {blue}\delta }+2u^{2}w{\color {crimson}\alpha }+2u^{2}v{\color {green}\beta }-v^{2}{\color {blue}\delta }+u^{4}-4v^{3}+10uvw=0$
The solution of that system then has to be entered in the already mentioned key to get the mentioned final form:
    $z=y^{4}+{\color {crimson}\alpha }y^{3}+{\color {green}\beta }y^{2}+{\color {orange}\gamma }y+{\color {blue}\delta }$
    $z^{5}+\lambda z-\mu =0$
The coefficients Lambda and My ofthe Bring Jerrard final form can be found out by doing a polynomial division of z^5 divided by the initial principal polynome and reading the resulting remainder rest. So a Bring Jerrard equation appears that contains only the quintic, the linear and the absolute term.

### Examples of solving the principal form

Along with the Abel Ruffini theorem the following equations are examples that can not be solved by elementary expressions, but can be reduced to the Bring Jerrard form by only using cubic radical elements. This shall be demonstrated here. To do this on the given principal quintics, we solve the equations for the coefficients of the cubic, quadratic and absolute term of the quartic Tschirnhaus key after the shown pattern. So this Tschirnhaus key can be determinded. By doing a polynomial division on the fifth power of the quartic Tschirnhaus transformation key and analyzing the remainder rest the coefficients of the mold can be determined too. And so the solutions of following given principal quintic equations can be computed:
    $y^{5}-5y^{2}+5y-5=0$ $u=5,\,v=5,\,w=5$ $20{\color {crimson}\alpha }-15{\color {green}\beta }=25$ $75{\color {crimson}\alpha }^{2}-75{\color {crimson}\alpha }+125{\color {crimson}\alpha }{\color {green}\beta }-50{\color {green}\beta }^{2}=75$ $15{\color {crimson}\alpha }+5{\color {blue}\delta }-20=0$ ${\color {crimson}\alpha }=-1,\,{\color {green}\beta }=-3,\,{\color {blue}\delta }=7$ $5{\color {orange}\gamma }^{3}+110{\color {orange}\gamma }^{2}+1100{\color {orange}\gamma }+3080=0$ ${\color {orange}\gamma }=-4{\sqrt {3}}\tanh {\bigl{\tfrac {1}{3}}\operatorname {artanh} ({\tfrac {4}{9}}{\sqrt {3}}){\bigr ]}-2$ $z=y^{4}-y^{3}-3y^{2}+{\bigl \-4{\sqrt {3}}\tanh {\bigl{\tfrac {1}{3}}\operatorname {artanh} ({\tfrac {4}{9}}{\sqrt {3}}){\bigr ]}-2{\bigr \}}y+7$ $z^{5}+14080{\bigl \6-{\sqrt {3}}\coth {\bigl{\tfrac {1}{3}}\operatorname {arcoth} ({\tfrac {7}{9}}{\sqrt {3}}){\bigr ]}{\bigr \}}z+11264{\bigl \2+15{\sqrt {3}}\tanh {\bigl{\tfrac {1}{3}}\operatorname {artanh} ({\tfrac {1}{27}}{\sqrt {3}}){\bigr ]}{\bigr \}}=0$
This is a further example for that algorithm:
    $y^{5}-5y^{2}+15y-12=0$ $u=5,\,v=15,\,w=12$ $60{\color {crimson}\alpha }-15{\color {green}\beta }=60$ $75{\color {crimson}\alpha }^{2}-225{\color {crimson}\alpha }+300{\color {crimson}\alpha }{\color {green}\beta }-150{\color {green}\beta }^{2}=-150$ $15{\color {crimson}\alpha }+5{\color {blue}\delta }-60=0$ ${\color {crimson}\alpha }=1,\,{\color {green}\beta }=0,\,{\color {blue}\delta }=9$ $5{\color {orange}\gamma }^{3}+135{\color {orange}\gamma }^{2}+990{\color {orange}\gamma }+2370=0$ ${\color {orange}\gamma }=-{\sqrt {10}}\coth {\bigl{\tfrac {1}{3}}\operatorname {arcoth} ({\tfrac {1}{2}}{\sqrt {10}}){\bigr ]}-4$ $z=y^{4}+y^{3}+{\bigl \-{\sqrt {10}}\coth {\bigl{\tfrac {1}{3}}\operatorname {arcoth} ({\tfrac {1}{2}}{\sqrt {10}}){\bigr ]}-4{\bigr \}}y+9$ $z^{5}+{\tfrac {3375}{151}}{\bigl \29{\sqrt {10}}\coth {\bigl{\tfrac {1}{3}}\operatorname {arcoth} ({\tfrac {2273}{145}}{\sqrt {10}}){\bigr ]}-16{\bigr \}}z+{\tfrac {675}{443}}{\bigl \1343{\sqrt {10}}\coth {\bigl{\tfrac {1}{3}}\operatorname {arcoth} ({\tfrac {45965}{1343}}{\sqrt {10}}){\bigr ]}-1145{\bigr \}}=0$

### Clues for creating the Moduli and Nomes

That Bring Jerrard equation can be solved by an elliptic Jacobi theta quotient that contains the fifth powers and the fifth roots of the corresponding elliptic nome in the theta function terms. For doing this, following elliptic modulus or numeric eccentricity and their Pythagorean counterparts and corresponding elliptic nome should be used in relation to Lambda and My after the essay Sulla risoluzione delle equazioni del quinto grado from Charles Hermite and Francesco Brioschi and the recipe on page 258 accurately:
    $f={\frac {5\mu }{4\lambda }}{\bigl{\frac {5}{\lambda }}{\bigr )}^{1/4}$
These are the elliptic moduli and thus the numeric eccentricities:
    $\varepsilon =\operatorname {ctlh} {\bigl{\tfrac {1}{2}}\operatorname {aclh} (f){\bigr ]}^{2}$ $Q=\exp {\bigl-\pi K({\sqrt {1-\varepsilon ^{2}}})\div K(\varepsilon ){\bigr ]}$ $\varepsilon ^{*}=\operatorname {tlh} {\bigl{\tfrac {1}{2}}\operatorname {aclh} (f){\bigr ]}^{2}$ $Q^{*}=\exp {\bigl-\pi K(\varepsilon )\div K({\sqrt {1-\varepsilon ^{2}}}){\bigr ]}$ $\varepsilon ^{*}={\sqrt {1-\varepsilon ^{2}}}$ $\ln(Q)\ln(Q^{*})=\pi ^{2}$
With the abbreviations ctlh abd tlh the Hyperbolic Lemniscatic functions are represented. The abbreviation aclh is the Hyperbolic Lemniscate Areacosine accurately.

## Literature

* "Polynomial Transformations of Tschirnhaus", Bring and Jerrard, ACM Sigsam Bulletin, Vol 37, No. 3, September 2003
* F. Brioschi, Sulla risoluzione delle equazioni del quinto grado: Hermite — Sur la résolution de l'Équation du cinquiéme degré Comptes rendus —. N. 11. Mars. 1858. 1. Dezember 1858, doi:10.1007/bf03197334
* Bruce and King, Beyond the Quartic Equation, Birkhäuser, 1996.