Editing Quartic function (section)

====Descartes' solution====
Descartes<ref>{{Citation|last = Descartes|first = René|author-link = René Descartes|title = [[La Géométrie|The Geometry of Rene Descartes with a facsimile of the first edition]]|isbn = 0-486-60068-8|publisher = [[Dover Publications|Dover]]|year = 1954|jfm = 51.0020.07|chapter = Book&nbsp;III: On the construction of solid and supersolid problems|orig-year = 1637}}</ref> introduced in 1637 the method of finding the roots of a quartic polynomial by factoring it into two quadratic ones. Let

:<math>
  \begin{align}
   x^4 + bx^3 + cx^2 + dx + e & = (x^2 + sx + t)(x^2 + ux + v) \\
   & = x^4 + (s + u)x^3 + (t + v + su)x^2 + (sv + tu)x + tv
  \end{align}
 </math>

By [[equating coefficients]], this results in the following system of equations:

:<math>
  \left\{\begin{array}{l}
   b = s + u \\
   c = t + v + su \\
   d = sv + tu \\
   e = tv
  \end{array}\right.
 </math>

This can be simplified by starting again with the [[#Converting to a depressed quartic|depressed quartic]] {{math|''y''<sup>4</sup> + ''py''<sup>2</sup> + ''qy'' + ''r''}}, which can be obtained by substituting {{math|''y'' − ''b''/4}} for {{math|''x''}}. Since the coefficient of {{math|''y''<sup>3</sup>}} is&nbsp;{{math|0}}, we get {{math|''s'' {{=}} −''u''}}, and:

:<math>
  \left\{\begin{array}{l}
   p + u^2 = t + v \\
   q = u (t - v) \\
   r = tv
  \end{array}\right.
 </math>

One can now eliminate both {{mvar|t}} and {{mvar|v}} by doing the following:
:<math>
  \begin{align}
   u^2(p + u^2)^2 - q^2 & = u^2(t + v)^2 - u^2(t - v)^2 \\
   & = u^2 [(t + v + (t - v))(t + v - (t - v))]\\
   & = u^2(2t)(2v) \\
   & = 4u^2tv \\
   & = 4u^2r
  \end{align}
 </math>

If we set {{math|''U'' {{=}} ''u''<sup>2</sup>}}, then solving this equation becomes finding the roots of the [[resolvent cubic]]

{{NumBlk|:|<math> U^3 + 2pU^2 + (p^2-4r)U - q^2,</math>|{{EquationRef|2}}}}

which is [[Cubic function#General solution to the cubic equation with real coefficients|done elsewhere]]. This resolvent cubic is equivalent to the resolvent cubic given above (equation (1a)), as can be seen by substituting U = 2m.

If {{math|''u''}} is a square root of a non-zero root of this resolvent (such a non-zero root exists except for the quartic {{math|''x''<sup>4</sup>}}, which is trivially factored),

:<math>
  \left\{\begin{array}{l}
  s = -u \\
 2t = p + u^2 + q/u \\
 2v = p + u^2 - q/u
  \end{array}\right.
 </math>

The symmetries in this solution are as follows. There are three roots of the cubic, corresponding to the three ways that a quartic can be factored into two quadratics, and choosing positive or negative values of {{mvar|u}} for the square root of {{mvar|U}} merely exchanges the two quadratics with one another.

The above solution shows that a quartic polynomial with rational coefficients and a zero coefficient on the cubic term is factorable into quadratics with rational coefficients if and only if either the resolvent cubic (''{{EquationNote|2}}'') has a non-zero root which is the square of a rational, or {{math|''p''<sup>2</sup> − 4''r''}} is the square of rational and {{math|''q'' {{=}} 0}}; this can readily be checked using the [[rational root test]].<ref name=Brookfield>{{cite journal |author=Brookfield, G. |title=Factoring quartic polynomials: A lost art |journal=[[Mathematics Magazine]] |volume=80 |issue=1 |year=2007 |pages=67–70|doi=10.1080/0025570X.2007.11953453 |s2cid=53375377 |url = https://www.maa.org/sites/default/files/Brookfield2007-103574.pdf}}</ref>