Problem:
Part of the graph of f(x)=ax3+bx2+cx+d is shown. What is b ?
Answer Choices:
A. −4
B. −2
C. 0
D. 2
E. 4
Solution:
We have
0=f(−1)=−a+b−c+d and 0=f(1)=a+b+c+d
so b+d=0. Also d=f(0)=2, so b=−2.
OR
The polynomial is divisible by (x+1)(x−1)=x2−1, its leading term is ax3, and its constant term is 2 , so
f(x)=(x2−1)(ax−2)=ax3−2x2−ax+2 and b=−2​.
The problems on this page are the property of the MAA's American Mathematics Competitions
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