How to show that $b_n(n) \neq 0$ for $b_n(x) = -\frac{16 x}{n} b_{n-1}(x) - \frac{48 x}{n} b_{n-2}(x)$? Lets define the recursive polynomial:
$b_0(x) = 1$
$b_1(x) = -16 x$
$b_n(x) = -\frac{16 x}{n} \left( b_{n-1}(x) + 3 b_{n-2}(x) \right) = -\frac{16 x}{n} b_{n-1}(x) - \frac{48 x}{n} b_{n-2}(x)$ ; with $x \in \mathbb{R}$ and $n \in \mathbb{N}$
Question 1: How can I show that $b_n(n) \neq 0$ ?
I can show that for the following recurrence relation $a_n \neq 0$.
$a_0 = 1$
$a_1 = -16$
$a_n = -16 \left( a_{n-1} + 3 a_{n-2} \right) = -16 a_{n-1} - 48 a_{n-2}$
The characteristic polynomial will be:
$\lambda^2 + 16\lambda + 48 = 0$
with the solutions $\lambda_1 = -12$ and $\lambda_2 = -4$.
Therefore I am able to express $a_n$ as:
$a_n = A_1 \lambda_1^n + A_2 \lambda_2^n = \frac{3}{2}(-12)^n - \frac{1}{2}(-4)^n = (-1)^n \frac{12^n - 4^n}{2} \neq 0$
My motivation was numerical evidence and that:
$b_n(n) = -16 \left( b_{n-1}(n) + 3 b_{n-2}(n) \right)$ which is like $a_n$ but
$b_{n-1}(n) = -16 \frac{n-1}{n} \left( b_{n-2}(n) + 3 b_{n-3}(n) \right)$ for example is only similar to $a_{n-1}$ in the long run when $n \to \infty$.
Question 2: Is it possible in any way to proof $b_n(n) \neq 0$ by using $a_n$ ?

If we fix $n$ and define $a_k=b_k(n)$, then indeed we obtain a recursion
$$a_0=1,\quad a_1=-16n,\quad a_k=-16a_{k-1}-48a_{k-2}. $$
From here we infer that
$$\tag1 a_k=A(-12)^k+B(-4)^k$$
for suitable $A,B$.
The desired result is that this makes $a_n\ne0$.
Clearly, $a_0\ne 0$. Also, when $n=1$, we have $a_1=-16\ne 0$, so that we need only consider the cases with $n\ge 2$.
From $(1)$ and the given values for $a_0,a_1$,  we  infer $$\tag2A+B=1,\quad -12A-4B=-16n,$$
Or more explicitely
$$ \tag3A=2n-\frac12,\quad B=\frac32-2n.$$
Note that $a_n=0$ implies
$$\frac BA=-3^n\le -9, $$whereas $(3)$ leads to 
$$\frac BA=\frac{\frac32-2n}{2n-\frac12}=-1+\frac1{2n-\frac12}>-1. $$
