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. $P(-1), P(0), P(1)$ $3k , k , 2k$ - Cubic $-2k , k$ - Quadratic $3k$ - Linear 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 $P(2)$ . $P(-2), P(-1), P(0), P(1), P(2)$ $8k , 3k , k , 2k , 6k$ - Cubic $-5k , -2k , k , 4k$ - Quadratic $3k , 3k , 3k$ - Linear 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
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2014 AMC 12B Problems/Problem 16

Contents

Problem

Solution

Solution 2

See also