Real polynomials and roots under certain restrictions Consider a real polynomial $r(x) = x^{n} + a_{n-1}x^{n-1} + ....a_{1}x - 1$. Assume that it has no root in $(-1,1)$, and that $r(-1)=0$. Then, which of the following are true?
$(\alpha)\lim_{x \to \infty} r(x) = \infty  $
$(\beta) r(1) = 0  $
$(\gamma)$ $ r(2) > 0$
Apparently, all three are true. My reasoning has been the following:
(a) Because the leading coefficient is positive, the first one is true.
(b) the constant in the polynomial is $-1$, and so there cannot be any root with absolute value greater than $1$, for otherwise there would be a root in $(-1,1)$, a contradiction to our assumption. With $r(-1)=0$, this implies the truth of the other two.
I found this problem interesting. Is there other approaches to this problem that we could undertake? As for example, by using other properties of polynomials which I am clearly oblivious to at the moment.
 A: Actually, both $(\beta)$ and $(\gamma)$ are false! Counterexamples must have complex roots of modulus $<1$. For example, with $c>1$, we can take
$$r(x) = (x^2+\tfrac1c)(x-c)(x+1)$$
as a family of counterexamples. Note that by varying $c$, $r(2)$ can be negative, positive, or zero.

To see the necessary condition on counterexamples, let $\lambda_i$ be the real roots of $r$ and let $\mu_j, \bar\mu_j$ be the conjugate pairs of complex roots. Then $r$ has constant term
$$\tag{$*$}
\prod_i(-\lambda_i)\prod_j|\mu_j|^2 \overset!=-1.
$$
It is not hard to see from this, that if $|\mu_j|\ge1$ for all $j$, then we must have $\lambda_i\in(-1,1]$ for some $i$.

By the way, I don't believe $-1$ being a root has any effect on the problem at all.

EDIT, assuming there are no roots with modulus $<1$. With this assumption, in order for $(*)$ to hold, we must have all roots on the unit circle (otherwise $|r(0)|>1$). So $r(0)$ becomes simply $(-1)^k$, where $k$ is the multiplicity of the root $1$. Since the result is negative, we must have $1$ as a root (and indeed with odd multiplicity).
For $(\gamma)$, consider a general monic real polynomial $f$ (with roots denoted as before), and a real number $t$. We have
$$
f(t) = \prod_i(t-\lambda_i)\prod_j|t-\mu_j|^2.
$$
If $f(t)\ne0$, then it follows that $f(t)>0$ if and only if there are an even number of real roots $>t$. So indeed $r(2)>0$ in our case (the number of real roots $>2$ being zero).
In fact,
$$
\mathrm{sign}(r(x))= \begin{cases}(-1)^{m+1} & x<1 \\ -1 & -1<x<1 \\ +1 & x>1 \\ 0 & x=\pm1 \end{cases}
$$
, where $m$ is the multiplicity of the root $-1$ (and $m$ is free to be any value).
A: Only (a) and (c) are true.
Your argument for (a) is perfectly valid.
(b) is false: consider $r_1(x) = x^3 +2x^2 -1$. %edit: this has a root in (-1,1) so maybe (b) is true as well
For (c): consider the factorized version of the polynom $r$ . as $a_n = 1$ we way write $r(x)$ as the product of $(x-b_i)$. As $a_0 = -1$, we have the product of $b_i$ being equal to $-1$.
Notice that $r(2) \ne 0$ otherwise on of the $b_i$ has an absolute value $<1$.
This argument actually holds for all $x >1$: $r(x) \ne 0$.
It follows that $r$ has no root beyond 1. As (a) is true, $r(x) > 0$ for $x>1$.
