1988 AHSME Problems/Problem 15

Problem

If $a$ and $b$ are integers such that $x^2 - x - 1$ is a factor of $ax^3 + bx^2 + 1$ , then $b$ is

$\textbf{(A)}\ -2\qquad \textbf{(B)}\ -1\qquad \textbf{(C)}\ 0\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 2$

Solution

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 $\boxed{\text{A}}.$

1988 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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1988 AHSME Problems/Problem 15

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