complex numbers - Can some cubic polynomial have two real roots?

In $p(x) = x^3-x^2$, both $0$ and $1$ are possible roots of the polynomial; both are real. I had read that a cubic polynomial has either all real roots or just one real root. It can't have two. What is the problem in this case?

Answer

In your case, $0$ is a double root: you should count it as two roots. In other words, the following statement holds:

If the roots are counted with their multiplicities, then every cubic polynomial in one variable with real coefficients either has exactly one real root or it has three real roots.