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.

  • 1
    $\begingroup$ $p(x)=x^3-\frac{2}{5}x^2-\frac{1}{40}, b=\frac{17}{40}, p\left(\frac{18}{40}\right)<0$ $\endgroup$ – 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


$n$ 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



$|r|\leq \left|a_1\right|+\left|a_2\right|\frac{1}{|r|}+\text{...}+|a_n|\frac{1}{\left|r|^{n-1}\right.}$

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 $p(x)<0$

  • $\begingroup$ why does it matter that they are odd? $\endgroup$ – user162089 May 6 '15 at 4:05
  • 1
    $\begingroup$ we need $n$ odd for $\lim_{x\rightarrow -\infty } p(x)=-\infty $ $\endgroup$ – Lozenges May 6 '15 at 12:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.