Showing that $m^3+4m+2=0$ has only one real root

I actually have to find the number of real roots of $m^3+4m+2=0$ for a conic sections question in which $m$ is the slope of the normal. The answer is that there is only one real value of $m$, and therefore only one normal.

How did they get this? How can I find out how many real roots such a cubic equation has without spending too much time on it?

The derivative of this function, $3m^2+4$, is always positive, so the function is always increasing. An increasing function on the real line cannot have more than one zero.

The function $f_1(m) = m^3$ is never decreasing. The function $f_2(m) = 2$ is constant. The function $f_3(m) = 4m$ is always increasing.

Add them together, you have a function $f(m) = f_1(m) + f_2(m) + f_3(m)$ that is always increasing. Once it passes through zero it cannot return to zero again.