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

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<cmath>P(2)+P(-2)=12k+2k=\boxed{\textbf{(E)}\ 14k}</cmath>
 
<cmath>P(2)+P(-2)=12k+2k=\boxed{\textbf{(E)}\ 14k}</cmath>
  
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== See also ==
 
{{AMC12 box|year=2014|ab=B|num-b=15|num-a=17}}
 
{{AMC12 box|year=2014|ab=B|num-b=15|num-a=17}}
 
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{{MAA Notice}}

Revision as of 13:24, 22 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 \[P(2)+P(-2)=12k+2k=\boxed{\textbf{(E)}\ 14k}\]

See also

2014 AMC 12B (ProblemsAnswer KeyResources)
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

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