# Need proof that a cubic equation has at least one real root

1. Apr 22, 2005

### lmamaths

Hi,

A cubic equation has at least one real root.
Can someone help me to prove this?

Thx!
LMA

2. Apr 22, 2005

### dextercioby

U need calculus.It's not difficult.Use the continuitiy on $\mathbb{R}$ and evaluate the 2 asymptotic limits.Then use the Rolle (IIRC) theorem...

Daniel.

3. Apr 22, 2005

### matt grime

I don't think you necessarily need that.

If w is a complex root of f(x) a real polynomial (ie one with real coeffs), then so is w* the conjugate of w, this means that the real poly (x-w)(x-w*) divides f(x) over the reals. Hence complex roots occur in conjugate pairs. A cubic has 3 (possibly complex) roots, so pairing up the complex ones (or any poly of odd degree) means there must be an odd number of unpaired real roots left.