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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 $p(x) >0$

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    $\begingroup$ Hint: Look at the behavior of $p(x)$ as $x\to -\infty$ and as $x\to\infty$. $\endgroup$ – rogerl Nov 2 '20 at 21:07
  • $\begingroup$ 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/… $\endgroup$ – player3236 Nov 2 '20 at 21:09
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    $\begingroup$ 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. $\endgroup$ – user2661923 Nov 2 '20 at 21:42
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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.

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