$q$ be a real polynomial of real variable $x$ of the form $q(x)=x^n+a_{n-1}x^{n-1}+....+a_1x-1 .\,\,$ I am stuck on the following problem:  

Let $q$ be a real polynomial of real variable $x$ of the form $q(x)=x^n+a_{n-1}x^{n-1}+....+a_1x-1 .\,\,$ Suppose $q$ has no roots in the open unit disc and $q(-1)=0.$
  Then which of the following options are correct?   
  
  
*
  
*$\lim_{x \to \infty}q(x)=\infty$    
  
*$q(3)=0$   
  
*$q(2) >0$  
  
*$q(1)=0.$  

I have also found the following result interesting but I am not sure it has anything to do with the current problem.  
 
I need some detailed clarification to get the desired result.
 A: (1) is correct because the leading coefficient is positive. Let $\alpha_1,\ldots,\alpha_n$ be the roots of $q$. We have $\prod\alpha_i=(-1)^{n+1}$. We see that if one of the roots had modulus $>1$ then another would have modulus $<1$, which contradicts the condition on the roots of $q$. Therefore all the roots lie on the unit circle so the roots split into real roots $\alpha_i=\pm 1$ and complex roots $\beta_j$ on the unit circle and since complex roots come in pairs it is of the form
$$
q(X)=(X-1)^k(X+1)^l\prod_{j\in J}|X-\beta_j|^2
$$
So $q(2)=3^l\prod_j|2-\beta_j|^2>0$. Also $q(3)$ can't be zero, which answers $(2)$. The formula $\prod\alpha_i=(-1)^{n+1}$ becomes $q(0)=(-1)^k=-1$, so $k>0$ and odd, which implies $q(1)=0$.
Summing up: (1), (3) and (4) are true and the opposite of (2) is true, i.e. $q(3)\neq 0$. Note also that every polynomial of the above form with $l>0$ satisfies the conditions, so this also gives all possible $q$'s.
Remark: This is not in contrast to Tomas answer. The difference here is that Tomas did not take into consideration the complex roots of $q$, in this answer I do.
A: (1) is true, because the leading coefficient of the highest monomial is positive.
All others are wrong, to prove this, take a look at the following polynomial:
$$(x-2)(x+1)(x^2+\frac{1}{2})=x^4-x^3-\frac{3}{2}x^2-\frac{1}{2}x-1$$
First, it is of the desired form. Further, its only roots are $2$ and $-1$ (since $x^2+\frac{1}{2}$ is irreducible). Hence, there is no root on the open unit disc and $q(-1)=0$, so it meets the preconditions.
But, $q(1)\neq 0, q(3)\neq 0$ and $q(2)=0$ so (2)-(4) are wrong.
