Prove that a polynomial function has a root Let $f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$
A polynomial function such that $n$ is odd positive integer and $a_n\ne0$
Prove that this function has a root.
I tried and eventually come to the conclusion that I need to prove that for $c \in R$ there exists $f(c) = 0$, but I'm not quite sure why, can someone please explain me the logic of this?
Furthermore, I know I need to use the intermediate value theorem, but I just don't know how to do it on this kind of expression, how do I solve this?
Thanks for your time!
 A: Hint: Consider $$\lim_{x \to \pm \infty} f(x).$$
Then, noting that polynomials are continuous everywhere, apply the Intermediate Value Theorem. 
Edit:
I will do one of the limits for you:
$$\lim_{x\to \infty} f(x) = \lim_{x\to \infty} x^n \left(a_n + \frac{a_{n-1}}{x}+ \cdots + \frac{a_1}{x^{n-1}} + \frac{a_0}{x^n} \right) = \infty.$$
A: Certainly you mean that the polynomial has real coefficients so by the intermediate value problem and since the limit of the polynomial at $\pm\infty$ is $\pm\infty$ or $\mp\infty$ we get the desired result. Otherwise and by the Gauss-D'Alembert theorem the polynomial has a complex root.
A: I think you have a polynomial with real coefficients and you mean real a zero.
According to the fundamental theorem of algebra this polynomial has exactly $n$ zeros in complex. Suppose all zeros are not real. Note that in a polynomial with real coefficients zeros are form a complex conjugates. Since $n$ is odd, we have a contradiction. Therefore  there exist at least one real zero.
