# Prove odd degree polynomials have roots. [duplicate]

Let $p:\mathbb{R}\rightarrow\mathbb{R}$ be a polynomial of odd degree. Prove that there is a solution of the equation $$p(x)=0, x\in\mathbb{R}$$

I am giving this question in an analysis textbook and the only machinery I have to work with is the sequential definition of continuity and the intermediate value theorem. Using only these ideas I am having difficulty coming up with a concrete proof.

• $\lim_{x\to+\infty} p(x) = -\lim_{x\to-\infty} p(x) = \pm\infty$. So we can pick $x$ sufficiently large that $p(x) < 0$ and $p(-x) > 0$. Then apply the IVT. – Xander Henderson Apr 12 '18 at 23:37
• What I am missing is a concrete proof that those limits are as stated using the sequential definition of convergence or divergence of a function. – Walt Apr 13 '18 at 2:19

Hint: if $p$ is a polynomial of odd degree, then $$\lim_{x\to +\infty}p(x)=\pm \infty, \quad \lim_{x\to -\infty}p(x)=\mp\infty$$
• The problem as I have stated it is that I need a concrete proof using sequences. Essentially if given a sequence call it $\{a_n\}$ of real values which diverges to infinity I have to show that $p(a_n)$ diverges as well for any odd degree polynomial. – Walt Apr 13 '18 at 2:19