2014 AMC 12B Problem 16

Attempt Problem 16 of the 2014 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2014 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

16.

Let PP be a cubic polynomial with P(0)=k,P(0) = k, P(1)=2k,P(1) = 2k, and P(−1)=3k.P(-1) = 3k. What is P(2)+P(−2)?P(2) + P(-2)?
00
kk
6k6k
7k7k
14k14k
Answer: E
Concepts:polynomialsystem of equationssymmetry (algebra)
Small hint:
Write P(x)=ax3+bx2+cx+kP(x) = ax^3 + bx^2 + cx + k using P(0)=kP(0)=k
Big hint:
Adding P(1)P(1) and P(−1)P(-1) isolates b,b, and P(2)+P(−2)=8b+2kP(2)+P(-2) = 8b + 2k
Solution:
Since P(0)=k,P(0) = k, write P(x)=ax3+bx2+cx+k.P(x) = ax^3 + bx^2 + cx + k.
Then P(1)=a+b+c+k=2kP(1) = a+b+c+k = 2k and P(−1)=−a+b−c+k=3k.P(-1) = -a+b-c+k = 3k. Adding these gives 2b+2k=5k,2b + 2k = 5k, so 2b=3k.2b = 3k.
The odd-power terms cancel in the sum: P(2)+P(−2)=(8a+4b+2c+k)+(−8a+4b−2c+k)=8b+2k. \begin{gathered} P(2)+P(-2) \\ = (8a+4b+2c+k) \\ {}+ (-8a+4b-2c+k) \\ = 8b + 2k. \end{gathered} Since 8b=4(2b)=12k,8b = 4(2b) = 12k, this equals 12k+2k=14k.12k + 2k = 14k.
Thus, the correct answer is E.
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