2014 AMC 12B Problem 16

Since $P(0) = k,$ write $P(x) = ax^3 + bx^2 + cx + k.$ Then $P(1) = a+b+c+k = 2k$ and $P(-1) = -a+b-c+k = 3k.$ Adding these gives $2b + 2k = 5k,$ so $2b = 3k.$ The odd-power terms cancel in the sum: $$\begin{gathered} P(2)+P(-2) \\ = (8a+4b+2c+k) \\ {}+ (-8a+4b-2c+k) \\ = 8b + 2k. \end{gathered}$$ Since $8b = 4(2b) = 12k,$ this equals $12k + 2k = 14k.$ Thus, the correct answer is E.