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Difference between revisions of "2003 AMC 12A Problems/Problem 21"

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== Problem 21 ==
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== Problem ==
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The graph of the polynomial
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The graph of the [[polynomial]]
  
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<math>P(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + e</math>
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<cmath>P(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + e</cmath>
  
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has five distinct <math>x</math>-intercepts, one of which is at <math>(0,0)</math>. Which of the following coefficients cannot be zero?
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has five distinct <math>x</math>-intercepts, one of which is at <math>(0,0)</math>. Which of the following [[coefficient]]s cannot be zero?
  
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<math>\textbf{(A)}\ a \qquad \textbf{(B)}\ b \qquad \textbf{(C)}\ c \qquad \textbf{(D)}\ d \qquad \textbf{(E)}\ e</math>
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<math>\text{(A)}\ a \qquad \text{(B)}\ b \qquad \text{(C)}\ c \qquad \text{(D)}\ d \qquad \text{(E)}\ e</math>
  
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__TOC__
 
== Solution ==
 
== Solution ==
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=== Solution 1 ===
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According to [[Vieta's formulas]], the sum of the roots of a 5th degree polynomial taken 4 at a time is <math>\frac{a_1}{a_5} = d</math> . Calling the roots <math>r_1, r_2, r_3, r_4, r_5</math> and letting <math>r_1 = 0</math> (our given zero at the origin), the only way to take four of the roots without taking <math>r_1</math> is <math>r_2r_3r_4r_5</math>.
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Since all of the other products of 4 roots include <math>r_1</math>, they are all equal to <math>0</math>. And since all of our roots are distinct, none of the terms in <math>r_2r_3r_4r_5</math> can be zero, meaning the entire expression is not zero. Therefore, <math>d</math> is a sum of zeros and a non-zero number, meaning it cannot be zero, so <math>\mathrm{(D)}</math>.
  
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According to Vieta's Formula, the sum of the roots of a 5th degree polynomial taken 4 at a time is <math>\frac{a_1}{a_5} = d</math> . Calling the roots <math>r_1, r_2, r_3, r_4, r_5</math> and letting <math>r_1 = 0</math> (our given zero at the origin), the only way to take four of the roots without taking <math>r_1</math> is <math>r_2r_3r_4r_5</math>.
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=== Solution 2 ===
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Since all of the other products of 4 roots include <math>r_1</math>, they are all equal to 0. And since all of our roots are distinct, none of the terms in <math>r_2r_3r_4r_5</math> can be zero, meaning the entire expression is not zero. Therefore, <math>d</math> is a sum of zeros and a non-zero number, meaning it cannot be zero. <math>\Rightarrow D</math>
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Clearly, since <math>(0,0)</math> is an intercept, <math>e</math> must be <math>0</math>. But if <math>d</math> was <math>0</math>, <math>x^2</math> would divide the polynomial, which means it would have a double root at <math>0</math>, which is impossible, since all five roots are distinct.
  
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== Solution 2 ==
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== See Also ==
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{{AMC12 box|year=2003|ab=A|num-b=20|num-a=22}}
  
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Clearly, since (0,0) is an intercept, e must be 0. But if d was 0, <math>x^2</math> would divide the polynomial, which means it would have a double root at 0, which is impossible, since all five roots are distinct.
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[[Category:Intermediate Algebra Problems]]
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== See Also ==
 
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*[[2003 AMC 12A Problems]]
 

Revision as of 15:30, 14 June 2008

Problem

The graph of the polynomial

\[P(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + e\]

has five distinct $x$-intercepts, one of which is at $(0,0)$. Which of the following coefficients cannot be zero?

$\text{(A)}\ a \qquad \text{(B)}\ b \qquad \text{(C)}\ c \qquad \text{(D)}\ d \qquad \text{(E)}\ e$

Solution

Solution 1

According to Vieta's formulas, the sum of the roots of a 5th degree polynomial taken 4 at a time is $\frac{a_1}{a_5} = d$ . Calling the roots $r_1, r_2, r_3, r_4, r_5$ and letting $r_1 = 0$ (our given zero at the origin), the only way to take four of the roots without taking $r_1$ is $r_2r_3r_4r_5$. Since all of the other products of 4 roots include $r_1$, they are all equal to $0$. And since all of our roots are distinct, none of the terms in $r_2r_3r_4r_5$ can be zero, meaning the entire expression is not zero. Therefore, $d$ is a sum of zeros and a non-zero number, meaning it cannot be zero, so $\mathrm{(D)}$.

Solution 2

Clearly, since $(0,0)$ is an intercept, $e$ must be $0$. But if $d$ was $0$, $x^2$ would divide the polynomial, which means it would have a double root at $0$, which is impossible, since all five roots are distinct.

See Also

2003 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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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