Difference between revisions of "2014 AMC 12B Problems/Problem 16"
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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) | + | <math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ k\qquad\textbf{(C)}\ 6k\qquad\textbf{(D)}\ 7k\qquad\textbf{(E)}\ 14k </math> |
==Solution== | ==Solution== | ||
Line 15: | Line 15: | ||
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> | ||
+ | |||
+ | 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 == | == 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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 23:34, 1 November 2024
Contents
Problem
Let be a cubic polynomial with , , and . What is ?
Solution
Let . Plugging in for , we find , and plugging in and for , we obtain the following equations: Adding these two equations together, we get If we plug in and in for , we find that Multiplying the third equation by and adding gives us our desired result, so
Solution 2
If we use Gregory's Triangle, the following happens:
Since this is cubic, the common difference is for the linear level so the string of s are infinite in each direction. If we put a on each side of the original , we can solve for and .
The above shows us that is and is so .
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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