1988 AHSME Problems/Problem 15

Revision as of 17:23, 26 February 2018 by Hapaxoromenon (talk | contribs) (Added a solution with explanation)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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


See also

1988 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS