Singapur

Expanding $(x+1)^n(x^2+bx+1)$, we obtain
$$x^{n+2} + (n+b)x^{n+1} + \sum_{k=0}^{n-2} \left[ \binom{n}{k+2} + b\binom{n}{k+1} + \binom{n}{k} \right] x^{n-k} + (n+b)x + 1.$$
We require $n > 2$ so that $n+b > 0$. We also require $\binom{n}{k+2} + b\binom{n}{k+1} + \binom{n}{k} > 0$ for all $k=0, \dots, n-2$ which is equivalent to require $H \equiv \frac{\binom{n}{k+2} + b\binom{n}{k+1} + \binom{n}{k}}{\binom{n}{k}} > 0$ for all $k=0, \dots, n-2$. Direct simplification gives
$$H = \frac{(2-b)k^2 + (b-2)(n-2)k + 2bn + n^2 - n + 2}{(k+2)(k+1)}.$$
The numerator of $H$ as a quadratic expression in $k$ has leading coefficient $(2-b) > 0$ and discriminant $[(b-2)(n-2)]^2 - 4(2-b)(2bn+n^2-n+2) = (n+2)(b-2)(bn+2b+2n)$. Since $b-2 < 0$, $H$ is positive if and only if $bn+2b+2n > 0$, which is equivalent to $n > -\frac{2b}{b+2}$.