Composite degrees *n* are better solved by *polynomial decomposition*. However, for , there is an alternative method. Thus, we’ll give two methods to solve the quartic. Given the depressed quartic,

Its ** resolvent cubic** is simply,

**Method 1**:

Given any root *y* of the cubic, then,

For convenience, one can choose the real root *y*. Notice that {} are the negative case of . The formula can easily be proven by squaring the variable *y*, forming the quadratic in {}, and eliminating *y* between them using resultants,

proving any root of the sextic (a cubic in ) will suffice.

**Method 2**:

Given the three roots {} of the resolvent cubic, then,

where signs are chosen such that,

[1]

*Proof*: Expanding,

using the formulas, we get,

From [1] and the properties of elementary symmetric polynomials, we find that it is in fact equal to,

A similar method was used to solve the octic in this post, though now one had to take square roots of its resolvent *septic*.