If $a$ and $b$ are integers such that $x^2 - x - 1$ is a factor of $ax^3 + bx^2 + 1$, then $b$ is
$\textbf{(A)}\ -2\qquad \textbf{(B)}\ -1\qquad \textbf{(C)}\ 0\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 2$
Using polynomial division, we find that the remainder is $(2a+b)x+(a+b+1)$, so for the condition to hold, we need this remainder to be $0$. This gives $2a+b=0$ and $a+b+1=0$, so $b=-2a$ and $a-2a+1=0 \implies a=1 \implies b=-2$, which is $\boxed{\text{A}}.$