Ans: B
Let $p(x) = (x^2-1)Q(x) +(ax+b)$, where $Q(x)$ is a polynomial, $a$ and $b$ are constants.
By the remainder theorem, we have
$\begin{array}{rcl}
p(-1) & = & -2 \\
[(-1)^2 -1]Q(-1) +a(-1)+b & = & -2 \\
-a+b & = & -2 \ldots \unicode{x2460}
\end{array}$
By the factor theorem, we have
$\begin{array}{rcl}
p(1) & = & 0 \\
[(1)^2 -1]Q(1) + a(1) +b & = & 0 \\
a+b & = & 0 \\
a & = & -b
\end{array}$
Sub. $a=-b$ into $\unicode{x2460}$, we have
$\begin{array}{rcl}
-(-b) +b & = & -2 \\
2b & = & -2 \\
b & = & -1
\end{array}$
Therefore, $a=1$.
Hence, the required remainder is $x-1$.
