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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...")
 
m (→‎Solution 2: what is gregory's triangle)
 
(7 intermediate revisions by 5 users not shown)
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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> ?
 
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==
 
==Solution==
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Multiplying the third equation by <math>4</math> and adding <math>2k</math> gives us our desired result, so  
 
Multiplying the third equation by <math>4</math> and adding <math>2k</math> gives us our desired result, so  
 
<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==
 +
If we use [[Gregory's Triangle]], the following happens:
 +
<cmath>P(-1), P(0), P(1)</cmath>
 +
<cmath>  3k ,  k  ,  2k </cmath>
 +
<cmath>    -2k , k      </cmath>
 +
<cmath>      3k        </cmath>
  
−
(Solution by kevin38017)
+
Since this is cubic, the common difference is <math>3k</math> for the linear level so the string of <math>3k</math>s are infinite in each direction.
 +
If we put a <math>3k</math> on each side of the original <math>3k</math>, we can solve for <math>P(-2)</math> and <math>P(2)</math>.
 +
 
 +
<cmath>P(-2), P(-1), P(0), P(1), P(2)</cmath>
 +
<cmath>  8k ,  3k ,  k  ,  2k ,  6k </cmath>
 +
<cmath>    -5k  , -2k ,  k  ,  4k    </cmath>
 +
<cmath>      3k  ,  3k ,  3k      </cmath>
 +
 
 +
The above shows us that <math>P(-2)</math> is <math>8k</math> and <math>P(2)</math> is <math>6k</math> so <math>8k+6k=14k</math>.
 +
 
 +
NOTE (not from author): The link you put for gregory's triangle doesn't work so please explain it in your post or find a resource that does work; there isn't much on google.
 +
 
 +
== See also ==
 +
{{AMC12 box|year=2014|ab=B|num-b=15|num-a=17}}
 +
{{MAA Notice}}

Latest revision as of 22:34, 1 November 2024

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 $8k+6k=14k$.

NOTE (not from author): The link you put for gregory's triangle doesn't work so please explain it in your post or find a resource that does work; there isn't much on google.

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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