Is this proof correct/complete?

Question

Let $f(x)=\sum\limits_{k=0}^{n}c_kx^k$ be a polynomial where $c_0$ and $c_n$ have different sings. Show $\exists x_0 \in \mathbb{R}$ such that $f(x_0)=0$.

My workings so far: Lets assume $c_0>0$ and thus $c_n<0$. If this is not the case we can simply look at $f^*(x)=-f(x)$ and $f^*(x)$ will satisfy this condition where $f^*(x_0)=0$ implies $f(x_0)=0$. By this assumption we know $f(0)>0$. Then for sufficiently large $x_l$ we have: $$|c_nx_l^n|>\left|\sum\limits_{k=0}^{n-1}c_kx_l^k\right|$$ Therefore, as $c_n<0$, it follows that: $$f(x_l)=\sum\limits_{k=0}^{n-1}c_kx_l^k-|c_n|x_l^n<0$$ Now because $f(x)$ is a polynomial and therefore continuous, we can apply the intermediate value theorem to conclude that $\exists x_0 \in [0,x_l]$ such that $f(x_0)=0$.

Answer 1

It is easier if you take the limit at $+\infty$.

Assume $c_0<0$ and $c_n>0$ first.

Then $f(0)=c_0<0$, while $\lim_{x\rightarrow +\infty} f(x)=\lim_{x\rightarrow +\infty}c_nx^n=+\infty$.

So we can find $x>0$ such that $f(x)>0$.

By the intermediate value theorem, there exists $x_0>0$ such that $f(x_0)=0$.

I let you do the case $c_0>0$ and $c_n<0$.