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

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.

using Descartes' Rule of Signs

It's plain that there will be no positive roots. And $f(-m)=-m^3-4m+2 \quad$ leads to one sign change, therefore 1 negative root.