1. Given that $a$, $b$, and $c$ are real numbers, and the polynomial $x^{3}+$ $a x^{2}+b x+c$ can be divided by $x^{2}+3 x-4$. Then $2 a-2 b$ $-c=$ $\qquad$ .
$=、 1.14$.
From the problem, -4 and 1 are roots of the equation $x^{3}+a x^{2}+b x+c=0$, which means
$$
\begin{array}{l}
16 a-4 b+c=64, \\
a+b+c=-1 .
\end{array}
$$
(1) $\times 6$ - (2) gives $2 a-2 b-c=14$.