2014 AMC 12B Problem 16

Problem 16 of 25IntermediateAlgebra

Let

P

be a cubic polynomial with

P(0) = k,

P(1) = 2k,

and

P(-1) = 3k.

What is

P(2) + P(-2)?

Show solution

Solution

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,

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.