If $c_{n} > 0$ then $\sum_{0}^{n}c_{k}x^{k} > 0$ for some $x \in \mathbb{R}$? Let $n \geq 1$ be an integer and let $c_{0}, \dots, c_{n} \in \mathbb{R}$. If $c_{n} > 0,$ is there necessarily an $x \in \mathbb{R}$ such that 
$$\sum_{0}^{n}c_{k}x^{k} > 0?$$
I just realized that for a while I had implicitly taken this for granted. However, when I would like to give a rigorous proof then I find it is not that obvious. 

Hint: Any $x > \frac{|c_{n-1}|+|c_{n-2}|+...+|c_0|}{c_n}+1$ works. 
This implies that $x>1$ and in particular 
$$c_nx^n > |c_{n-1}|x^{n-1}+|c_{n-2}|x^{n-1}+...+|c_0|x^{n-1} \geq |c_{n-1}|x^{n-1}+|c_{n-2}|x^{n-2}+...+|c_0| \\ \geq -c_{n-1}x^{n-1}-c_{n-2}x^{n-2}-...-c_0  $$
