Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

What bound is there on the roots of a given polynomial, in terms of its degree and coefficients?

Consider the polynomial $p(x) = 3x^7 – 5x^3 + 42$. Would you not agree, without doing any calculation, that one million ($10^6$) cannot be a root? It just wouldn’t be in accord with the smallness of the coefficients and the well-behavedness of polynomials. And yet I don’t recall ever having encountered anything in the literature that gave a bound on the absolute value of the roots of a polynomial in terms of the degree and coefficients of the polynomial, but I’m pretty sure such must exist, and that I simply missed it, and so I’m tagging this a a reference-request.

By the way, during my post of a question, every time, it seems, there are stray graphics on the screen, as if from someone else's question or answer. Is this happening to anyone else?

share|cite|improve this question
This Wikipedia article looks helpful. I'd also google "polynomial root bounds" or some such and look at the material. – anon Aug 2 '11 at 1:14
You may find this article useful: Bounding the Roots of Polynomials.. It's over $\mathbb{C}[x]$, but I guess similar results can be shown for fields of characteristic zero. – user2468 Aug 2 '11 at 1:51
@M.S.: what does it mean to bound roots in a field without an absolute value? – Qiaochu Yuan Aug 2 '11 at 3:56
@anon: You are right. After I posted my question, I realized that I had forgotten to do the elementary thing of googling for the topic. When I say that I have never seen something in the literature, what I mean is that many years ago when I used haunt the stacks of the library perusing mathematics books, I never saw it. However, in this case, the material on the web is so rich, that I am grateful for Qiaochu Yuan’s cutting through the mass of material and giving me a spot-on answer, which I am therefore up-voting and accepting. (By the way, I love your poetry:) – Mike Jones Aug 3 '11 at 18:28
up vote 12 down vote accepted

Here's an elementary bound. Let $p(x) = x^n - a_{n-1} x^{n-1} - ... - a_0$. If $|x| \ge 1$ then $m \ge n$ implies $|x|^m \ge |x|^n$, hence if $p(x) = 0$ then

$$|x|^n = \left| a_{n-1} x^{n-1} + ... + a_0 \right| \le |x|^{n-1} \left( |a_{n-1}| + ... + |a_0| \right)$$


$$|x| \le \text{max}(1, |a_{n-1}| + ... + |a_0|).$$

In your case we can do much better because of the zero coefficients: we get

$$|x|^7 \le \text{max}(1, \frac{47}{3} |x|^3)$$

hence $|x|^4 \le \frac{47}{3} < 16$, or $|x| < 2$.

A technique useful in special cases is the Gauss-Lucas theorem, and a technique useful in general cases is Rouche's theorem.

share|cite|improve this answer
Totally cool! Thanks! I love MSE! – Mike Jones Aug 3 '11 at 18:29

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.