Editing Quartic function (section)

===Simpler cases===
====Reducible quartics====
Consider the general quartic
:<math>Q(x) = a_4x^4+a_3x^3+a_2x^2+a_1x+a_0.</math>
It is [[irreducible polynomial|reducible]] if {{math|''Q''(''x'') {{=}} ''R''(''x'')×''S''(''x'')}}, where {{math|''R''(''x'')}} and {{math|''S''(''x'')}} are non-constant polynomials with [[rational number|rational]] coefficients (or more generally with coefficients in the same [[field (mathematics)|field]] as the coefficients of {{math|''Q''(''x'')}}). Such a factorization will take one of two forms:

:<math>Q(x) = (x-x_1)(b_3x^3+b_2x^2+b_1x+b_0)</math>
or
:<math>Q(x) = (c_2x^2+c_1x+c_0)(d_2x^2+d_1x+d_0).</math>
In either case, the roots of {{math|''Q''(''x'')}} are the roots of the factors, which may be computed using the formulas for the roots of a [[quadratic function]] or [[cubic function]].

Detecting the existence of such factorizations can be done [[Resolvent cubic#Factoring quartic polynomials|using the resolvent cubic of {{math|''Q''(''x'')}}]]. It turns out that:
* if we are working over {{math|'''R'''}} (that is, if coefficients are restricted to be real numbers) (or, more generally, over some [[real closed field]]) then there is always such a factorization;
* if we are working over {{math|'''Q'''}} (that is, if coefficients are restricted to be rational numbers) then there is an algorithm to determine whether or not {{math|''Q''(''x'')}} is reducible and, if it is, how to express it as a product of polynomials of smaller degree.

In fact, several methods of solving quartic equations ([[Quartic function#Ferrari's solution|Ferrari's method]], [[Quartic function#Descartes' solution|Descartes' method]], and, to a lesser extent, [[Quartic function#Euler's solution|Euler's method]]) are based upon finding such factorizations.

====Biquadratic equation====
If {{math|''a''<sub>3</sub> {{=}} ''a''<sub>1</sub> {{=}} 0}} then the function 
:<math>
Q(x) = a_4x^4+a_2x^2+a_0
</math> 
is called a '''biquadratic function'''; equating it to zero defines a '''biquadratic equation''', which is easy to solve as follows

Let the auxiliary variable {{math|''z'' {{=}} ''x''<sup>2</sup>}}.
Then {{math|''Q''(''x'')}} becomes a [[Quadratic function|quadratic]] {{math|''q''}} in {{math|''z''}}: {{math|''q''(''z'') {{=}} ''a''<sub>4</sub>''z''<sup>2</sup> + ''a''<sub>2</sub>''z'' + ''a''<sub>0</sub>}}. Let {{math|''z''<sub>+</sub>}} and {{math|''z''<sub>−</sub>}} be the roots of {{math|''q''(''z'')}}. Then the roots of the quartic {{math|''Q''(''x'')}} are
:<math>
\begin{align}
x_1&=+\sqrt{z_+},
\\
x_2&=-\sqrt{z_+},
\\
x_3&=+\sqrt{z_-},
\\
x_4&=-\sqrt{z_-}.
\end{align}
</math>

==== Quasi-palindromic equation ====
The polynomial
: <math>P(x)=a_0x^4+a_1x^3+a_2x^2+a_1 m x+a_0 m^2</math>
is almost [[Reciprocal polynomial#Palindromic polynomial|palindromic]], as {{math|''P''(''mx'') {{=}} {{sfrac|''x''<sup>4</sup>|''m''<sup>2</sup>}}''P''({{sfrac|''m''|''x''}})}} (it is palindromic if {{math|''m'' {{=}} 1}}). The change of variables {{math|''z'' {{=}} ''x'' + {{sfrac|''m''|''x''}}}} in {{math|{{sfrac|''P''(''x'')|''x''<sup>2</sup>}} {{=}} 0}} produces the [[quadratic equation]]  {{math|''a''<sub>0</sub>''z''<sup>2</sup> + ''a''<sub>1</sub>''z'' + ''a''<sub>2</sub> − 2''ma''<sub>0</sub> {{=}} 0}}. Since {{math|''x''<sup>2</sup> − ''xz'' + ''m'' {{=}} 0}}, the quartic equation {{math|''P''(''x'') {{=}} 0}} may be solved by applying the [[quadratic formula]] twice.