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 $8k+6k=14k$ .

2014 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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2014 AMC 12B Problems/Problem 16

Contents

Problem

Solution

Solution 2

See also