What conditions do this polynomial satisfy? Be p(x) a monic polynomial with real and positive coefficients, degree n >= 2 and with n real roots. What of this alternatives are right?
$(a) p(2) < 2(2^{n−1} + 1)\\ (b) p(1) < 3 \\(c) p(1) ≥ 2^{n} \\(d) p(3) < 3(2^{n−1} − 2)$
I have been seeking for consequences of this conditions, and how does them impact in the polynomial, but i am finding anything. Particularly i think the conditions n real roots and monic polynomial is the key to answer this question, but i don't know what to do with this properties...
Would appreciate any help.
 A: Let $p(x) = x^n + a_{n-1}x_{n-1} + \dots + a_1x + a_0$ with $a_i > 0,\ a_i\in\mathbb{R}$. Since all the coefficients are positive, for any $x>0$, we have $p(x)>0$. Hence, all the $n$ real roots are negative. Let $-\alpha_i$ be the $i$-th root of $p(x)$. Note that here $\alpha_i$ will be strictly positive. Then, we can rewrite $p(x)$ as
$$ p(x) = (x + \alpha_1)\dots(x + \alpha_n).$$
Now,
(1). $p(1) = \prod_{i=1}^n (1+\alpha_i)$ tells us that $p(1) > 1$.
(2). $p(2) = \prod_{i=1}^n (2+\alpha_i)$ tells us that $p(2) > 2^n$.
(3). $p(3) = \prod_{i=1}^n (3+\alpha_i)$ tells us that $p(3) > 3^n$.
One can choose suitable $\alpha_i$'s so that $p(1) > 3$. Hence, (b) is false. For example, $\alpha_1 = 100$.
Next, consider option (d).
$p(3) < 3(2^{n-1} - 2) < 3(2^{n-1}-2) + 6 < 3\cdot2^{n-1} < 3\cdot3^{n-1} < 3^n$ contradicts (3) above. Hence, (d) is false.
For option (c), set $\alpha_i = 0.5$ for all $1\leq i\leq n$. Then, $p(1) < 1.5^n < 2^n$. Hence, (c) is also false.
But even option (a) seems to be a bad bound because I can simply choose my $\alpha_i$'s to be equal to $4$ which will make the $p(2) = 6^n > 2(2^{n-1} + 2)$ for $n\geq 2$.
Choosing a set of $\alpha_i$'s means there exists a polynomial with real coefficients and $n$-real roots such that the above conditions do not hold. This is a general way to check. Maybe you need to check the question. Might have messed up some inequality. Hope this helps.
