Do cubics always have one real root? I've seen a few conflicting pieces of information online. 
So far, I know that with real coefficients there will always be one real root. But how about with complex coefficients? 
At very least could you give me a counterexample? A cubic with no real roots.
 A: One of the best things you can remember is that over a field (like the reals or complex numbers) roots come from linear factors.  Use this to build your own examples:  $f(z) =(z-i)^3$.  If you want three distinct complex roots, do something like $f(z) = (z-i)(z+i)(z-2i)$.
A: As you already know, a cubic with real coefficients always has at least one real root, so there is no counterexample of a cubic with real coefficients with no real roots.
A cubic with complex coefficients with no real roots is easy to find; take $x^3+i$.
A: Over the complex numbers, every polynomial factors into roots. So we can take any cubic and write it as $a(x-u)(x-v)(x-w)$ where $u, v, w$ are the roots (they don't need to be distict) and $a$ is the leading coefficient. This lets us form polynomials with only complex roots such as $(x-i)^3$. 
However, if all the original coefficients of the polynomial are real, and $c$ is a complex root, then its conjugate $\bar{c}$ must also be a root: complex roots to polynomials with real coeffecients must come in pairs. This is called the complex conjugate root theorem.
This means that a polynomial with real coefficients and odd degree will always have at least one real root, which answers the case for cubics. A quadratic with negative discriminant on the other hand has two, conjugate complex roots.
