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Let $p$ be a real polynomial of the real variable $x$ of the form $$p(x)=x^n+a_{n-1}x^{n-1}+...+a_1x-1$$Suppose that $p$ has no roots in the open unit disc and $p(-1)=0$.Then which is/are true?

$1$.$p(1)=0$.

$2$.$\displaystyle{\lim_{x\to \infty}}p(x)=\infty$

$3$.$p(2)>0$

$4$.$p(3)=0$

product of the roots of the polynomial is $-1$. and it is a monic polynomial.If it has a root whose modulus is greater than one then it must have a root whose modulus is less than one which gives a contradicton. So it has roots $1$ or $-1$. since one root is $-1$ and product of the roots is $-1$ then it has a root $-1$.So 1 is true and 4 is false.obviously 2 is true.
I am not sure about 3.
since p has no root outside $[-1,1]$ so it does not change sign in $[1,\infty)$.So $p(2)>0$ or $p(2)<0$.But then I can not proceed.can I get some help?

poton
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  • Roots with imaginary parts occur in conjugate pairs. The formula $p(x) = \prod_k (x-r_k)$ consists of two parts, terms with real roots, and the others. If you look at $p(2)$ and use the fact that $|r_k| = 1$, you can answer 3. – copper.hat Jun 26 '13 at 17:28
  • Am I right for the others? – poton Jun 26 '13 at 17:36
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    The product of the roots is $\pm 1$, not $-1$, specifically, the product of the roots is $(-1)^{n+1}$. That doesn't affect the rest of your argument, however. – Thomas Andrews Jun 26 '13 at 17:48
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    Should $2$ read $\lim_{x \to \infty}p(x)$? If so, what is the $\lim_{x \to \infty} x^{n}$? – preferred_anon Jun 26 '13 at 18:13

1 Answers1

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then we can write the polynomial as p(x)=(x-1)(x+1) with some powers say k and r respectively with the condition k+r=n . then substitute x=2 in that we get 3.

lavanya nathan
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