Problem:
Let P be a cubic polynomial with P(0)=k,P(1)=2k, and P(−1)=3k. What is P(2)+P(−2) ?
Answer Choices:
A. 0
B. k
C. 6k
D. 7k
E. 14k
Solution:
Because P(0)=k, it follows that P(x)=ax3+bx2+cx+k. Thus P(1)=a+b+c+k=2k and P(−1)=−a+b−c+k=3k. Adding these equations gives 2b=3k. Hence
P(2)+P(−2)=(8a+4b+2c+k)+(−8a+4b−2c+k)=8b+2k=12k+2k=14k.
OR
Let (P(−2),P(−1),P(0),P(1),P(2))=(r,3k,k,2k,s). The sequence of first differences of consecutive values is (3k−r,−2k,k,s−2k), the sequence of second differences is (r−5k,3k,s−3k), and the sequence of third differences is (8k− r,s−6k). Because P is a cubic polynomial, the third differences are equal, so P(−2)+P(2)=r+s=14k.
The problems on this page are the property of the MAA's American Mathematics Competitions