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Difference between revisions of "2000 AMC 12 Problems/Problem 22"

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The [[graph]] below shows a portion of the [[curve]] defined by the quartic [[polynomial]] <math>P(x) = x^4 + ax^3 + bx^2 + cx + d</math>. Which of the following is the smallest?
 
The [[graph]] below shows a portion of the [[curve]] defined by the quartic [[polynomial]] <math>P(x) = x^4 + ax^3 + bx^2 + cx + d</math>. Which of the following is the smallest?
  
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<math>\text{(A)}\ P(-1)\\
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[[Image:2000_12_AMC-22.png]]
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\text{(B)}\ \text{The\ product\ of\ the\ zeros\ of\ } P\\
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\text{(C)}\ \text{The\ product\ of\ the\ non-real\ zeros\ of\ } P \\
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<math>\textbf{(A)}\ P(-1)\\
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\text{(D)}\ \text{The\ product\ of\ the\ coefficients\ of\ } P \\
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\textbf{(B)}\ \text{The\ product\ of\ the\ zeros\ of\ } P\\
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\text{(E)}\ \text{The\ product\ of\ the\ real\ zeros\ of\ } P</math>
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\textbf{(C)}\ \text{The\ product\ of\ the\ non-real\ zeros\ of\ } P \\
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\textbf{(D)}\ \text{The\ sum\ of\ the\ coefficients\ of\ } P \\
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\textbf{(E)}\ \text{The\ sum\ of\ the\ real\ zeros\ of\ } P</math>
  
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{{image}}
 
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<!-- The graph is a quartic, I can’t figure out how to reproduce it without the exact coefficients, but I’ll try -->
 
 
== Solution ==
 
== Solution ==
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We note that there are no more zeros of this polynomial, as there already have been three turns in the curve. We approximate each of the above expressions:
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Note that there are 3 maxima/minima. Hence we know that the rest of the graph is greater than 10. We approximate each of the above expressions:
 
# According to the graph, <math>P(-1) > 4</math>
 
# According to the graph, <math>P(-1) > 4</math>
 
# The product of the roots is <math>d</math> by [[Vieta’s formulas]]. Also, <math>d = P(0) > 5</math> according to the graph.
 
# The product of the roots is <math>d</math> by [[Vieta’s formulas]]. Also, <math>d = P(0) > 5</math> according to the graph.
 
# The product of the real roots is <math>>5</math>, and the total product is <math><6</math> (from above), so the product of the non-real roots is <math>< \frac{6}{5}</math>.
 
# The product of the real roots is <math>>5</math>, and the total product is <math><6</math> (from above), so the product of the non-real roots is <math>< \frac{6}{5}</math>.
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# The sum of the coefficients is <math>P(1) > 1.5</math>
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# The sum of the coefficients is <math>P(1) > 2.5</math>
 
# The sum of the real roots is <math>> 5</math>.
 
# The sum of the real roots is <math>> 5</math>.
  
 
Clearly <math>\mathrm{(C)}</math> is the smallest.
 
Clearly <math>\mathrm{(C)}</math> is the smallest.
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== See also ==
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== Video Solution ==
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{{AMC12 box|year=2000|num-b=20|num-a=22}}
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https://youtu.be/MMIEbkGu-k8
 +
 
 +
== See Also ==
 +
{{AMC12 box|year=2000|num-b=21|num-a=23}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 14:37, 5 June 2022

Problem

The graph below shows a portion of the curve defined by the quartic polynomial $P(x) = x^4 + ax^3 + bx^2 + cx + d$. Which of the following is the smallest?

2000 12 AMC-22.png

$\textbf{(A)}\ P(-1)\\ \textbf{(B)}\ \text{The\ product\ of\ the\ zeros\ of\ } P\\ \textbf{(C)}\ \text{The\ product\ of\ the\ non-real\ zeros\ of\ } P \\ \textbf{(D)}\ \text{The\ sum\ of\ the\ coefficients\ of\ } P \\ \textbf{(E)}\ \text{The\ sum\ of\ the\ real\ zeros\ of\ } P$

Solution

Note that there are 3 maxima/minima. Hence we know that the rest of the graph is greater than 10. We approximate each of the above expressions:

  1. According to the graph, $P(-1) > 4$
  2. The product of the roots is $d$ by Vieta’s formulas. Also, $d = P(0) > 5$ according to the graph.
  3. The product of the real roots is $>5$, and the total product is $<6$ (from above), so the product of the non-real roots is $< \frac{6}{5}$.
  4. The sum of the coefficients is $P(1) > 2.5$
  5. The sum of the real roots is $> 5$.

Clearly $\mathrm{(C)}$ is the smallest.

Video Solution

https://youtu.be/MMIEbkGu-k8

See Also

2000 AMC 12 (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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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