In mathematics, a quartic equation is one which can be expressed as a quartic function equaling zero. The general form of a quartic equation is
a x 4 + b x 3 + c x 2 + d x + e = 0 {\displaystyle ax^{4}+bx^{3}+cx^{2}+dx+e=0\,}
where a ≠ 0. The quartic is the highest order polynomial equation that can be solved by radicals in the general case.
History Lodovico Ferrari is attributed with the discovery of the solution to the quartic in 1540, but since this solution, like all algebraic solutions of the quartic, requires the solution of a cubic to be found, it could not be published immediately. The solution of the quartic was published together with that of the cubic by Ferrari's mentor Gerolamo Cardano in the book Ars Magna (1545). The proof that this was the highest order general polynomial for which such solutions could be found was first given in the Abel–Ruffini theorem in 1824, proving that all attempts at solving the higher order polynomials would be futile. The notes left by Évariste Galois before his death in a duel in 1832 later led to an elegant complete theory of the roots of polynomials, of which this theorem was one result.
Special case solutions Consider a quartic equation expressed in the form a 0 x 4 + a 1 x 3 + a 2 x 2 + a 3 x + a 4 = 0 {\displaystyle a_{0}x^{4}+a_{1}x^{3}+a_{2}x^{2}+a_{3}x+a_{4}=0} : There exists a general formula for finding the roots to quartic equations, provided the coefficient of the leading term is non-zero. However, since the general method is quite complex and susceptible to errors in execution, it is better to apply one of the special cases listed below if possible.
Degenerate case If the constant term a4 = 0, then one of the roots is x = 0, and the other roots can be found by dividing by x, and solving the resulting cubic equation,
a 0 x 3 + a 1 x 2 + a 2 x + a 3 = 0. {\displaystyle a_{0}x^{3}+a_{1}x^{2}+a_{2}x+a_{3}=0.\,}
Evident roots: 1 and −1 and −k Call our quartic polynomial Q(x). Since 1 raised to any power is 1,
Q ( 1 ) = a 0 + a 1 + a 2 + a 3 + a 4 . {\displaystyle Q(1)=a_{0}+a_{1}+a_{2}+a_{3}+a_{4}\ .}
Thus if a 0 + a 1 + a 2 + a 3 + a 4 = 0 , {\displaystyle \ a_{0}+a_{1}+a_{2}+a_{3}+a_{4}=0\ ,} Q(1) = 0 and so x = 1 is a root of Q(x). It can similarly be shown that if a 0 + a 2 + a 4 = a 1 + a 3 , {\displaystyle \ a_{0}+a_{2}+a_{4}=a_{1}+a_{3}\ ,} x = −1 is a root. In either case the full quartic can then be divided by the factor (x − 1) or (x + 1) respectively yielding a new cubic polynomial, which can be solved to find the quartic's other roots. If a 1 = a 0 k , {\displaystyle \ a_{1}=a_{0}k\ ,} a 2 = 0 {\displaystyle \ a_{2}=0\ } and a 4 = a 3 k , {\displaystyle \ a_{4}=a_{3}k\ ,} then x = − k {\displaystyle \ x=-k\ } is a root of the equation. The full quartic can then be factorized this way:
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