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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?

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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
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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

1 Answer 1

up vote 10 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

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

hence

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

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.

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Totally cool! Thanks! I love MSE! –  Mike Jones Aug 3 '11 at 18:29

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