💎 Exploring Quartic 

Ferrari's Resolvent Cubic 

A depressed quartic:
y⁴ + p y² + q y + r = 0

Note that q ≠ 0. Otherwise, the equation is simply a quadratic in y² and can be solved directly.

We want to factor it into (y² + ...) (y² + ...) = 0.

Letting e, f, E, F be some real coefficients to be found.

(y² + e y + f) (y² + E y + F) = 0

Expanding:
y⁴ + (E + e) y³ + (F + f + e E) y² + (e F + f E) y + f F = 0

Since there is no y³ term, E + e = 0, E = - e.

Therefore,
y⁴ + (F + f - e²) y² + (e F - e f) y + F f = 0

Comparing coefficients:
p = F + f - e²
q = e (F - f)
r = F f

From p = F + f - e²:
(F + f) = p + e²

From q = e (F - f):
(F - f) = q / e

Interestingly, (F + f)² = (F - f)² + 4 F f

Therefore,
(p + e²)² = (q / e)² + 4 r

Expanding,
p² + 2 p e² + (e²)² = q² / e² + 4 r

Let z = e²:
p² + 2 p z + z² = q² / z + 4 r
p² z + 2 p z² + z³ = q² + 4 r z
z³ + 2 p z² + (p² - 4 r) z - q² = 0

Since e is real, then z = e² must be real.
Since q ≠ 0 and (F - f) = q / e, then e ≠ 0.
Thus, z = e² > 0.

Therefore, we need to solve this cubic equation for a positive value of z. If there are multiple possible values of z, either one is fine. They would end up with different factorization of the same quartic and ultimately produce the same four roots y, just arranged differently.

With the same logic, after finding a suitable z, e = +√ z. Choosing - √ would only swap values of F and f to some other values but final four roots of y stay the same anyway.

Having e, calculate:

• F = (p + e² + q / e) / 2

• f = (p + e² - q / e) / 2

• E = - e

Then, easily solve (y² + e y + f) (y² + E y + F) = 0 with 2 quadratics. Phew.