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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.

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    $\begingroup$ $p(x)=x^3-\frac{2}{5}x^2-\frac{1}{40}, b=\frac{17}{40}, p\left(\frac{18}{40}\right)<0$ $\endgroup$ – Lozenges May 3 '15 at 15:30
  • $\begingroup$ @Lozenges you neeed to take absolute value $\endgroup$ – user162089 May 4 '15 at 2:31
  • $\begingroup$ The counterexample I gave is to show that your value of $b$ is not correct. Try $b=\max \left(1,\left|a_1\right|+\text{...}+|a_n|\right)$ $\endgroup$ – Lozenges May 4 '15 at 9:00
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$p(x)=x^n+a_1x^{n-1}+\text{...}+a_{n-1}x+a_n$

$n$ 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 $p(x)<0$

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  • $\begingroup$ why does it matter that they are odd? $\endgroup$ – user162089 May 6 '15 at 4:05
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    $\begingroup$ we need $n$ odd for $\lim_{x\rightarrow -\infty } p(x)=-\infty $ $\endgroup$ – Lozenges May 6 '15 at 12:25

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