This is a question from a past paper which I have no solution to.
Let $p(x)=x^n + a_{1}x^{n-1}+\cdots+a_{n-1}x+a_{n}, n\geq 1$ be a polynomial of dgree n and let $b=1+|a_{1}|+\cdots +|a_{n-1}|+|a_{n}|$.
Suppose that $n$ is odd. Show that $p(x)>0$ for every $x\geq b$ and $p(x)<0$ for every $x\leq -b$.
I have not made any meaning progress so far, nor have I been able to make any interesting observations. I considered expanding but it was obviously too tedious and quickly appeared to be a stupid idea. I considered applying some kind of MVT for $p(x)$ or the intergral of p(x) with 0 as the constant of integration, but nothing interesting so far. any thoughts?
edit: a minor mistake in what the value of $b$ should taken to be.
$p(x)=x^n+a_1x^{n-1}+\text{...}+a_{n-1}x+a_n$
$n$ odd, let $c=max(1,b)$ where $b=\left|a_1\right|+\text{...}+|a_n|$
First show that all the real roots of $p(x)$ are in the interval $(-c,c)$
Let $r$ be a nonzero root
$r^n=-a_1r^{n-1}-a_2r^{n-2}-\text{...}-a_{n-1}r-a_n$
$r=-a_1-a_2\frac{1}{r}-\text{...}-a_{n-1}\frac{1}{r^{n-2}}-a_n\frac{1}{r^{n-1}}$
$|r|\leq \left|a_1\right|+\left|a_2\right|\frac{1}{|r|}+\text{...}+|a_n|\frac{1}{\left|r|^{n-1}\right.}$
so if $|r|>1$ then $|r|\leq b$
This proves that $|r|\leq c$
Next, we show that $x>c$ implies $p(x)>0$ for otherwise $p(x)<0$ and since $\lim_{x\to \infty } p(x)=+\infty $, by $IVT$ there would be a root of $p(x)$ greater than $c$. a contradiction
similarly, if $p(x)>0$ for some $x<-c$ and since $\lim_{x\to -\infty } p(x)=-\infty $ ($n$ is odd) again by $IVT$ there would be a root of $p(x)$ less than $-c$. contradiction
This shows that $x>c$ implies $p(x)>0$ and $x<-c$ implies $p(x)<0$