2014 AMC 12B Problem 16 — Solution | LIVE by Po-Shen Loh

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2014 AMC 12B Problem 16

Below is the professionally curated solution for Problem 16 of the 2014 AMC 12B, 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.

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Concepts:polynomial system of equations symmetry (algebra)

Difficulty rating: 1950

16.

Let $P$ be a cubic polynomial with $P(0) = k,$ $P(1) = 2k,$ and $P(-1) = 3k.$ What is $P(2) + P(-2)?$

$0$

$k$

$6k$

$7k$

$14k$

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 $2b + 2k = 5k,$ so $2b = 3k.$

The odd-power terms cancel in the sum: $P(2)+P(-2) = (8a+4b+2c+k) + (-8a+4b-2c+k) = 8b + 2k.$ Since $8b = 4(2b) = 12k,$ this equals $12k + 2k = 14k.$

Thus, the correct answer is E.

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