Solve ax³ + bx² + cx + d = 0 for all roots.
First, divide by a to get x³ + px² + qx + r = 0. Substitute x = t − p/3 to get the depressed cubic t³ + mt + n = 0. Then use Cardano's formula with discriminant analysis.
The discriminant Δ = −4m³ − 27n² of the depressed cubic determines root types. If Δ > 0, three distinct real roots. If Δ = 0, repeated roots. If Δ < 0, one real and two complex roots.
For t³ + mt + n = 0: let D = (n/2)² + (m/3)³. Then t = ∛(−n/2 + √D) + ∛(−n/2 − √D). This was discovered by Cardano in 1545 and was a breakthrough in algebra.
Yes. A cubic with real coefficients can have either 3 real roots or 1 real root and 2 complex conjugate roots. It always has at least one real root (since odd-degree polynomials cross the x-axis).