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Given a polynomial of degree $n$, and the possible coefficients of polynomials are restricted to an interval for each of the degree. Is there a way to estimate number of roots of this polynomial in a given interval $[x_1,x_2]$.

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By "polynomial of order n", do you mean degree n? That is, your polynomial is $a_n x^n + \dots + a_1 x + a_0$? –  Dimitrije Kostic Nov 24 '11 at 12:37
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Given an interval $[x_1,x_2]$ you can check if $f(x_1) f(x_2) <0$ and in this case you can deduce there exists at least one root of the polynomial in the interval. If $f'(x)<0$ or $f'(x) >0$ for every $x \in [x_1,x_2]$ then the root is unique. –  user17090 Nov 24 '11 at 13:31
    
To Dimitrije Kostic: Ya order I meant degree. –  user16409 Nov 24 '11 at 14:21
    
Do you mean to impose different conditions on each $a_i$, or the same condition on all of the $a_i$? –  Dylan Yott Jun 12 '13 at 3:59
    
@DylanYott Question isn't mine, but I read "an interval for each of the degree[s]" as having an interval for each $a_i$. –  ˈjuː.zɚ79365 Jun 12 '13 at 7:30
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2 Answers

You cannot expect any good estimate near repeated roots. For instance, let $n$ be even and let $a_n$ lie in $[1-\epsilon,1+\epsilon]$ while all other $a_i$ lie in $[-\epsilon,\epsilon]$. This can have anywhere from 0 to 2$n$ real roots (for instance, consider $x^n+(\epsilon/2)x^k$).

Sufficiently far from repeated roots, you can use normal methods of rootfinding. You can try them even if you're not sure about repeated roots. For instance, intermediate value theorem still works if $f(x_1)$ and $f(x_2)$ are completely positive and completely negative intervals, respectively. The rule of signs works if all intervals are strictly positive or strictly negative. But there is a form of chaos near any repeated root.

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Given certain restrictions on coefficients you may be able to apply Sturm's theorem or produce a variant thereof (if you are not familiar with Sturm's theorem, it has a similar flavor to Descartes' Rule of Signs). A reference that you may find useful here is Prasolov's "Polynomials."

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