Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to prove that all the roots of the polynomial $f(x)=a_o+a_1 x+\cdots+ x^n$ with real coefficients belong to the interval $ [-M, M] $, with $\displaystyle{M=1+\sum_{k=0}^{n}|a_k|}$

share|cite|improve this question
Hi Jonas. I honestly thought the exercise but not to use: ( – Roiner Segura Cubero Dec 7 '12 at 3:56
Induction tried but did not succeed – Roiner Segura Cubero Dec 7 '12 at 4:00
As written the claim is false, since the roots of $\,x^2+1\in\Bbb R[x]\,$ do not belong to the interval $\,[-3,3]\,$ ...Are there any other conditions on the polynomial? – DonAntonio Dec 7 '12 at 4:14
Excuse me Don Antonio. The exercise refers to the real roots of the polynomial – Roiner Segura Cubero Dec 7 '12 at 4:19
At first I did not know about that .. but lately if I accept all the answers to my questions. – Roiner Segura Cubero Dec 7 '12 at 4:20

But let me give you a slightly different (and in a sense, stronger) formulation that is true.

Claim: if $P(x) = x^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0$, then a real root $r$ of $P$ satisfies $|r| \leq 1 + \max|a_i|$.

Suppose WLOG that $|r|>1$. Then $P(r) = 0$ means that $r^n = -a_0 - a_1r - \ldots - a_{n-1}r^{n-1}$. Let's call $M = \max|a_i|$. Then since $|r^n| \leq |a_0| + |a_1||r| + \ldots + |a_{n-1}||r^{n-1}|$, we have that $|r|^n \leq M(1 + |r| + \ldots + |r|^{n-1})$. This is a finite geometric series, and writing this out gives that $|x|-1 \leq M(1 - \frac{1}{|r|^n}) \leq M$, as was to be proven. $\diamondsuit$

share|cite|improve this answer

If $|x|\ge1$ and $f(x)=0$ then $$|x^n|=|-a_0-a_1x-\cdots-a_{n-1}x^{n-1}|\le|x|^{n-1}\sum|a_i|$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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