Difference between revisions of "2002 AMC 12A Problems/Problem 25"

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The sum of the coefficients of <math>P</math> and of <math>Q</math> will be equal, so <math>P(1) = Q(1)</math>. The only answer choice with an intersection between the two graphs at <math>x = 1</math> is <math>(B)</math>. (The polynomials in the graph are <math>P(x) = 2x^4-3x^2-3x-4</math> and <math>Q(x) = -2x^4-2x^2-2x-2</math>.)
 
The sum of the coefficients of <math>P</math> and of <math>Q</math> will be equal, so <math>P(1) = Q(1)</math>. The only answer choice with an intersection between the two graphs at <math>x = 1</math> is <math>(B)</math>. (The polynomials in the graph are <math>P(x) = 2x^4-3x^2-3x-4</math> and <math>Q(x) = -2x^4-2x^2-2x-2</math>.)
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==Solution 2==
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We know every coefficient is equal, so we get <math>ax^n + ... + ax + a = 0</math> which equals <math>x^n + ... + x + 1 = 0</math>. We see apparently that x cannot be positive, for it would yield a number greater than zero for <math>Q(x)</math>. We look at the zeros of the answer choices. A, C, D, and E have a positive zero, which eliminates them. B is the answer.
  
 
==See Also==
 
==See Also==

Revision as of 00:30, 24 July 2016

Problem

The nonzero coefficients of a polynomial $P$ with real coefficients are all replaced by their mean to form a polynomial $Q$. Which of the following could be a graph of $y = P(x)$ and $y = Q(x)$ over the interval $-4\leq x \leq 4$?

2002AMC12A25.png

Solution

The sum of the coefficients of $P$ and of $Q$ will be equal, so $P(1) = Q(1)$. The only answer choice with an intersection between the two graphs at $x = 1$ is $(B)$. (The polynomials in the graph are $P(x) = 2x^4-3x^2-3x-4$ and $Q(x) = -2x^4-2x^2-2x-2$.)

Solution 2

We know every coefficient is equal, so we get $ax^n + ... + ax + a = 0$ which equals $x^n + ... + x + 1 = 0$. We see apparently that x cannot be positive, for it would yield a number greater than zero for $Q(x)$. We look at the zeros of the answer choices. A, C, D, and E have a positive zero, which eliminates them. B is the answer.

See Also

2002 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
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All AMC 12 Problems and Solutions

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