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

(→‎Solution 2)
(→‎Solution 2)
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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>
 
==Solution 2==
 
==Solution 2==
−
If we use Gregory's Triangle, the following happens.
+
If we use [[Gregory's Triangle]], the following happens:
 
<cmath>P(-1), P(0), P(1)</cmath>
 
<cmath>P(-1), P(0), P(1)</cmath>
 
<cmath>  3k ,  k  ,  2k </cmath>
 
<cmath>  3k ,  k  ,  2k </cmath>

Revision as of 20:57, 13 March 2023

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 $8k+6k=14k$.

See also

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

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