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

Revision as of 10:15, 3 March 2015

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 $P(2)+P(-2)=12k+2k=\boxed{\textbf{(E)}\ 14k}$

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.

@@ Line 3: / Line 3: @@
 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> \textbf{(A)}\ 0\qquad\textbf{(B)}\ k\qquad\textbf{(C)}\ 6k\qquad\textbf{(D)}}\ 7k\qquad\textbf{(E)}\ 14k </math>
+<math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ k\qquad\textbf{(C)}\ 6k\qquad\textbf{(D)}\ 7k\qquad\textbf{(E)}\ 14k </math>
 ==Solution==

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

Revision as of 10:15, 3 March 2015

Problem

Solution

See also