4. For what least integer $n$ are all solutions of the equation

$$
x^{3}-(5 n-9) x^{2}+\left(6 n^{2}-31 n-106\right) x-6(n-8)(n+2)=0 \text { greater than }-1 \text { ? }
$$
Solution. Rewrite the equation as

$$
x^{3}-(5 n-9) x^{2}+\left(6 n^{2}-31 n-106\right) x-6 n^{2}+36 n+96=0 .
$$

Notice that $x_{1}=1>-1$ is a solution for all $n$ (the sum of the coefficients of the polynomial on the left side of the equation is zero for all $n$). Divide the equation by $x-1$:

$$
x^{2}-5(n-2) x+6\left(n^{2}-6 n-16\right)=0
$$

The roots of this equation are: $x_{2}=3(n+2), x_{3}=2(n-8)$. We seek when these solutions are greater than -1:

$$
\left\{\begin{array} { l } 
{ 3 ( n + 2 ) > - 1 } \\
{ 2 ( n - 8 ) > - 1 }
\end{array} \rightarrow \left\{\begin{array}{l}
n>-\frac{7}{3}, n \in Z \\
n>\frac{15}{2}, n \in Z
\end{array} \rightarrow n \geq 8\right.\right.
$$

Since $n$ is an integer, then $n \geq 8, n \in Z$.

Answer: $n_{\min }=8$.