Problem:
If a and b are integers such that x2−x−1 is a factor of ax3+bx2+1, then b is
Answer Choices:
A. −2
B. −1
C. 0
D. 1
E. 2
Solution:
By long division, one finds that x2−x−1 divides ax3+bx2+1 with quotient ax+(a+b) and remainder (2a+b)x+(a+b+1). But x2−x−1 is a factor of ax3+bx2+1, so the remainder is 0. In other words,
2a+b=0a+b=−1
Solving, one obtains a=1,b=−2.
OR
Since x2−x−1 is a factor of ax3+bx2+1, the quotient must be ax−1 (why?). Thus
ax3+bx2+1=(ax−1)(x2−x−1)=ax3+(−a−1)x2+(1−a)x+1
Equating the coefficients of x2 on the left and right, and then the coefficients of x, we obtain
b=−a−1,0=1−a
Hence a=1 and b=−2.