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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?

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    $\begingroup$ 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. $\endgroup$ – harshit54 Jan 2 '19 at 12:45
  • $\begingroup$ 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 $0$. $\endgroup$ – Dietrich Burde Jan 2 '19 at 13:46
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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.

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Note: if you have complex coefficients instead, the statement isn’t true.

As an example, consider $f(z) = (z-1)(z+1)(z-i)$. $f$ has two real roots, but one complex root.

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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.

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