← Complex numbers
7 Polynomial equations with complex roots
This is the question that closes almost every paper: one complex root of a real cubic is given, and you must find the rest — plus an unknown real coefficient — in one pass.The method, every time
- A polynomial with real coefficients has complex roots in conjugate pairs: if a+bi is a root, a−bi is automatically a second root — you never have to find it separately.
- (z − (a+bi))(z − (a−bi)) expands to z² − 2az + (a²+b²), a real QUADRATIC factor you can write down immediately.
- Divide the cubic by that quadratic factor (or compare coefficients term by term) to find the unknown constant and the real third root together — do not try to guess the third root first.
- The remainder theorem, f(a) = remainder, opens almost every paper and is pure real algebra: substitute the given root of x − a into the polynomial and solve for the unknown coefficient.
- A discriminant condition (b² − 4ac = 0 for equal roots, < 0 for no real roots, > 0 for two real roots) turns "how many real roots" into one inequality or equation to solve for a parameter.