Problem:
Part of the graph of f(x)=ax3+bx2+cx+d is shown. What is b ?
Answer Choices:
A. β4
B. β2
C. 0
D. 2
E. 4
Solution:
We have
0=f(β1)=βa+bβc+d and 0=f(1)=a+b+c+d
so b+d=0. Also d=f(0)=2, so b=β2.
OR
The polynomial is divisible by (x+1)(xβ1)=x2β1, its leading term is ax3, and its constant term is 2 , so
f(x)=(x2β1)(axβ2)=ax3β2x2βax+2 and b=β2β.
The problems on this page are the property of the MAA's American Mathematics Competitions