-1. A.
From the fact that $\pm 1$ are roots of the given polynomial, we have
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ 1 + a + b + c = 0 , } \\
{ - 1 + a - b + c = 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=-c, \\
b=-1
\end{array}\right.\right. \\
\Rightarrow \frac{a b+3 a}{c}=-2 .
\end{array}
$$
1. Given $a$, $b$, $c$ are three non-zero real numbers, and $x^{2}-1$ is a factor of the polynomial $x^{3}+a x^{2}+b x+c$. Then the value of $\frac{a b+3 a}{c}$ is ( ).
(A) -2
(B) -1
(C) 1
(D) 2