Prove that the odd polynomial equation $p(x)=0$ has at least one real solution (use intermediate value theorem) [duplicate]

Let $$a_0,a_1,....a_n$$ be real numbers and $$p(x) = a_nx^n+a_{n-1}x^{n-1}+...a_1x+a_0$$ for $$x \in R$$

I am trying to prove that the odd polynomial equation $$p(x)=0$$ has at least one real solution by using the intermediate value theorem.

Here's how I started it.

We can assume without loss of generality that $$a_n=1$$ and can write $$p(x)= x^n(1+\frac{a_{n-1}}{x}+\frac{a_{n-2}}{x^2}+...\frac{a_0}{x_n})$$

How can I show that there exists $$M>0$$ such that $$x\leq -M$$ implies $$p(x) <0$$
and $$M>0$$ such that $$x\geq M$$ implies $$p(x) >0$$

• Hint: Look at the behavior of $p(x)$ as $x\to -\infty$ and as $x\to\infty$. – rogerl Nov 2 '20 at 21:07
• When $M$ is very large, all the $\frac {a_i}{x^j}$ tend to zero, so eventually the number in the parenthesis becomes positive. Also see math.stackexchange.com/questions/689575/… – player3236 Nov 2 '20 at 21:09
• I am giving this response as a comment, rather than answer for three reasons: (1) You specifically asked about the use of the IVT (2) I suspect without being sure, that if a polynomial is of even degree, then it can not be an odd polynomial (3) I suspect without being sure, that when the polynomial has all real coefficients, that the complex roots must come in conjugate pairs. Assuming so, since a polynomial of odd degree must have an odd number of roots, one of these roots must be real. – user2661923 Nov 2 '20 at 21:42

Hint: consider the limits $$\lim_{x\to \infty}p(x)$$ and $$\lim_{x\to -\infty}p(x)$$. If these are large and small enough you might be able to find such $$M$$’s quite easily.