Editing Quartic function (section)

====Solving by Lagrange resolvent====
The [[symmetric group]] {{math|''S''<sub>4</sub>}} on four elements has the [[Klein four-group]] as a [[normal subgroup]]. This suggests using a '''{{visible anchor|resolvent cubic}}''' whose roots may be variously described as a discrete Fourier transform or a [[Hadamard matrix]] transform of the roots; see [[Lagrange resolvents]] for the general method. Denote by {{math|''x<sub>i</sub>''}}, for {{math|''i''}} from&nbsp;{{math|0}} to&nbsp;{{math|3}}, the four roots of {{math|''x''<sup>4</sup> + ''bx''<sup>3</sup> + ''cx''<sup>2</sup> + ''dx'' + ''e''}}. If we set

: <math> \begin{align}
s_0 &= \tfrac12(x_0 + x_1 + x_2 + x_3), \\[4pt]
s_1 &= \tfrac12(x_0 - x_1 + x_2 - x_3), \\[4pt]
s_2 &= \tfrac12(x_0 + x_1 - x_2 - x_3), \\[4pt]
s_3 &= \tfrac12(x_0 - x_1 - x_2 + x_3),
\end{align}</math>

then since the transformation is an [[Involution (mathematics)|involution]] we may express the roots in terms of the four {{math|''s<sub>i</sub>''}} in exactly the same way. Since we know the value {{math|''s''<sub>0</sub> {{=}} −{{sfrac|''b''|2}}}}, we only need the values for {{math|''s''<sub>1</sub>}}, {{math|''s''<sub>2</sub>}} and {{math|''s''<sub>3</sub>}}. These are the roots of the polynomial

:<math>(s^2 - {s_1}^2)(s^2-{s_2}^2)(s^2-{s_3}^2).</math>

Substituting the {{math|''s<sub>i</sub>''}} by their values in term of the {{math|''x<sub>i</sub>''}}, this polynomial may be expanded in a polynomial in {{math|''s''}} whose coefficients are [[symmetric polynomial]]s in the {{math|''x<sub>i</sub>''}}. By the [[fundamental theorem of symmetric polynomials]], these coefficients may be expressed as polynomials in the coefficients of the monic quartic. If, for simplification, we suppose that the quartic is depressed, that is {{math|''b'' {{=}} 0}}, this results in the polynomial
{{NumBlk|:|<math> s^6+2cs^4+(c^2-4e)s^2-d^2 </math>|{{EquationRef|3}}}}
This polynomial is of degree six, but only of degree three in {{math|''s''<sup>2</sup>}}, and so the corresponding equation is solvable by the method described in the article about [[cubic function]]. By substituting the roots in the expression of the {{math|''x<sub>i</sub>''}} in terms of the {{math|''s<sub>i</sub>''}}, we obtain expression for the roots. In fact we obtain, apparently, several expressions, depending on the numbering of the roots of the cubic polynomial and of the signs given to their square roots. All these different expressions may be deduced from one of them by simply changing the numbering of the {{math|''x<sub>i</sub>''}}.

These expressions are unnecessarily complicated, involving the [[root of unity|cubic roots of unity]], which can be avoided as follows. If {{math|''s''}} is any non-zero root of (''{{EquationNote|3}}''), and if we set

:<math>\begin{align}
F_1(x) & = x^2 + sx + \frac{c}{2} + \frac{s^2}{2} - \frac{d}{2s} \\
F_2(x) & = x^2 - sx + \frac{c}{2} + \frac{s^2}{2} + \frac{d}{2s}
\end{align}</math>

then

:<math>F_1(x)\times F_2(x) = x^4 + cx^2 + dx + e.</math>

We therefore can solve the quartic by solving for {{math|''s''}} and then solving for the roots of the two factors using the [[quadratic formula]].

This gives exactly the same formula for the roots as the one provided by [[Quartic function#Descartes' solution|Descartes' method]].