Proving $1+x+\frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}>0$. I've been asked to prove that $1+x+\frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}>0$. Proving that $1+x+\frac{x^2}{2!}>0$ is simple as you just need to consider the discriminant and show that it's less that 0. Showing $1+x+\frac{x^2}{2!} + \frac{x^3}{3!}=0$ only has 1 solution is to just consider the derivative and using the previous proof that $1+x+\frac{x^2}{2!}>0$ has no zero (so it is always increasing).
With this information, how would you be able to prove $1+x+\frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}>0$? I've tried to show that the global minimum (the lone solution to $1+x+\frac{x^2}{2!} + \frac{x^3}{3!}=0$) gives a positive answer, however I have not been successful. How can I show this?

As you say, $1+x+x^2/2+x^3/6=0$ has a unique solution, (with $x<0$). For this $x$,
$$1+x+\frac{x^2}2+\frac{x^3}{3!}+\frac{x^4}{4!}=\frac{x^4}{4!}>0.$$
