Ans: C
Since $g(x)$ is divisible by $x – 1$, by the factor theorem, we have
Since $g(x)$ is divisible by $x – 1$, by the factor theorem, we have
$\begin{array}{rcl}
g(1) & = & 0 \\
(1)^8 + a(1)^7 + b & = & 0 \\
b & = & -a – 1
\end{array}$
By the remainder theorem, the required remainder
$\begin{array}{cl}
= & g(-1) \\
= & (-1)^8 + a(-1)^7 + b \\
= & 1 -a – a – 1 \\
= & – 2a
\end{array}$
