# 2014 AMC 12B Problems/Problem 16

## Problem

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

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ k\qquad\textbf{(C)}\ 6k\qquad\textbf{(D)}\ 7k\qquad\textbf{(E)}\ 14k$

## Solution

Let $P(x) = Ax^3+Bx^2 + Cx+D$. Plugging in $0$ for $x$, we find $D=k$, and plugging in $1$ and $-1$ for $x$, we obtain the following equations: $$A+B+C+k=2k$$ $$-A+B-C+k=3k$$ Adding these two equations together, we get $$2B=3k$$ If we plug in $2$ and $-2$ in for $x$, we find that $$P(2)+P(-2) = 8A+4B+2C+k+(-8A+4B-2C+k)=8B+2k$$ Multiplying the third equation by $4$ and adding $2k$ gives us our desired result, so $$P(2)+P(-2)=12k+2k=\boxed{\textbf{(E)}\ 14k}$$

## Solution 2

If we use Gregory's Triangle, the following happens: $$P(-1), P(0), P(1)$$ $$3k , k , 2k$$ $$-2k , k$$ $$3k$$

Since this is cubic, the common difference is $3k$ for the linear level so the string of $3k$s are infinite in each direction. If we put a $3k$ on each side of the original $3k$, we can solve for $P(-2)$ and $P(2)$.

$$P(-2), P(-1), P(0), P(1), P(2)$$ $$8k , 3k , k , 2k , 6k$$ $$-5k , -2k , k , 4k$$ $$3k , 3k , 3k$$

The above shows us that $P(-2)$ is $8k$ and $P(2)$ is $6k$ so $8k+6k=14k$.

## See also

 2014 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 15 Followed byProblem 17 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions

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