Let $f(z)$ be a degree $3$ complex polynomial with real coefficients.
... tells you precisely that you can think of $f$ as a function $f\colon\mathbb R\to\mathbb R$ and can use intermediate value theorem.
After you obtain one real root $\alpha$, divide $f$ by $x-\alpha$ to get a degree $2$ polynomial. I assume you know how to prove how many roots it has.
Of course if
$f(x) \in \Bbb R[x], \tag 1$
then complex roots come together in conjugate pairs, since if
$f(x) = \displaystyle \sum_0^n f_ix^i, \; f_i \in \Bbb R, \tag 2$
and
$f(\rho) = 0, \tag 3$
we have
$\displaystyle \sum_0^n f_i \bar \rho^i = \sum_0^n \bar f_i \bar \rho^i = \overline{\sum_0^n f_i \rho} = \overline{f(\rho)} = 0, \tag 4$
that is,
$f(\rho) = 0 \Longleftrightarrow f(\bar \rho) = 0. \tag 5$
As for the existence of such pairs of roots, this is a much more difficult issue; as pointed out, the case of quadratic $f$, $n = 2$, has been completely solved. In the case $n = 3$ that is
$f(x) = ax^3 + bx^2 + cx + d = 0, \tag 6$
there is in fact a known criterion for the existence of complex zeroes; indeed, if the discriminant
$\Delta = 18abcd - 4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2 < 0, \tag 7$
then (6) has a complex conjugate pair of roots. Of course, being of odd degree, $f(x)$ always has at least one real zero. See this wikipedia page on cubic polynomials for a detailed explanation. Similar reults are known in the quartic case $n = 4$, but the formulas are so complicated I won't copy them here; just click the link to get the full story.
Our general knowledge is exhausted by the cases $n \le 4$; for $n \ge 5$, one must resort to a variety of specialized methods.
Best Answer
What do you mean with complex root? If you consider real numbers as particular complex numbers ($i.e.$ complex numbers with zero-imaginary part) then every polynomial of the form $p(z)=(z-z_0)^n$ where $z_0$ is a real number has exactly one complex root.
But if you mean a complex number which is not real; then this is impossible. Indeed, let $z_0=a+ib$ be any complex number; then the only polynomials which have $z_0$ as unique complex solution are $p(z)=(z-z_0)^n$ where $n$ is a non-zero natural number. But clearly, in this case, your polynomial has not real coefficients.