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?

• I think the statement is that if a cubic polynomial has three distinct roots, then either all 3 are real or only one of them is real and the other two are complex conjugates of each other. – harshit54 Jan 2 '19 at 12:45
• The title is not good. A trivial answer would be: Yes, take $f(x)=x^3$. It has $3$ real roots, $0$, $0$ and again $0$. – Dietrich Burde Jan 2 '19 at 13:46

In your case, $$0$$ is a double root: you should count it as two roots. In other words, the following statement holds:
As an example, consider $$f(z) = (z-1)(z+1)(z-i)$$. $$f$$ has two real roots, but one complex root.
The point is that complex conjugation $${\Bbb C}\rightarrow {\Bbb C}:z\mapsto \bar z$$, where $$\bar z = a-ib$$ if $$z=a+ib$$, is a (ring) automorphism. It follows that if you have a polynomial (with real coefficients in your case) $$f(x)$$ with $$f(z)=0$$ for some $$z\in{\Bbb C}$$, then $$f(\bar z) = \overline {f(z)} = \bar 0 = 0$$. Thus complex roots always occur in pairs: $$(z,\bar z)$$.
So actually a cubic polynomial with real coefficients can only have 1 or 3 real roots, but not 2. If if would have 2 real roots and 1 complex root $$z$$, then $$\bar z$$ would also be a root and so (as argumented above) $$z=\bar z$$ would be real.