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The following holds:

Let $P(x)$ be a polynomial in one variable $x$ of degree $3$ with complex coefficients such that

a) $$ P(-1)=P(1)=0 $$

Then

b) the formal derivative $P^{'}(x)$ has a root in the region

$$ R = \lbrace z : \mid Re(z)\mid < 1 \rbrace $$

of the complex plane.

Want to know if it is possible to prove this known fact by an appropriate application of Rouche's theorem.

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1  
What would a "no" answer to this question look like? The only thing I can think of would be to do some reverse math and find a reasonable theory where Rouch$\acute{e}$'s theorem is provable but this statement is not. However this statement can be proved easily by integrating the real part of $P′(x)$ and Rouch$\acute{e}$'s theorem is normally proved with some integration (correct?) so I suspect it would be hard to separate the two in this way. –  Amit Kumar Gupta Sep 8 '11 at 23:06
3  
How does it follow from Gauss-Lucas? There is a third root of $P$ that may be outside the region $R$. Actually, though, it follows from the fact that $P$ is of the form $P(x) = (a x + b )(x^2 - 1)$ that the product of the roots of $P'$ is $-1/3$. So at least one of those roots has absolute value $\le 1/\sqrt{3}$. –  Robert Israel Sep 9 '11 at 2:06
    
[previous comment, making risibly wrong statement -- as pointed out by Robert Israel -- deleted] –  Yemon Choi Sep 10 '11 at 1:01
    
dear amit: can you eventually ellaborate on your comment (line 3 of it mainly) (e.g. transforming it with more work in an answer) this will be useful in order to compare with robert's solution. –  Luis H Gallardo Sep 26 '11 at 20:09
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