2. Given that $m$, $n$, and $p$ are real numbers. If $x-1$ and $x+4$ are both factors of the polynomial $x^{3}+m x^{2}+n x+p$, then
$$
2 m-2 n-p+86=
$$
$\qquad$.
2. 100 .

From the divisibility property of polynomials, we know
$$
\left\{\begin{array} { l } 
{ 1 + m + n + p = 0 , } \\
{ - 6 4 + 1 6 m - 4 n + p = 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
p=12-4 m, \\
n=3 m-13 .
\end{array}\right.\right.
$$

Therefore, $2 m-2 n-p+86$
$$
\begin{array}{l}
=2 m-2(3 m-13)-(12-4 m)+86 \\
=100 .
\end{array}
$$