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Difference between revisions of "2014 AMC 12B Problems/Problem 16"

(Created page with "==Problem== Let <math>P</math> be a cubic polynomial with <math>P(0) = k</math>, <math>P(1) = 2k</math>, and <math>P(-1) = 3k</math>. What is <math>P(2) + P(-2)</math> ? <math...")
(No difference)

Revision as of 19:35, 20 February 2014

Problem

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

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ k\qquad\textbf{(C)}\ 6k\qquad\textbf{(D)}}\ 7k\qquad\textbf{(E)}\ 14k$ (Error compiling LaTeX. Unknown error_msg)

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 by kevin38017)

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