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

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== Problem 21 ==
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== Problem ==
The graph of the polynomial
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The graph of the [[polynomial]]
  
<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>
  
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?
  
<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>
  
 
== Solution ==
 
== Solution ==
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=== Solution 1 ===
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Let the roots be <math>r_1=0, r_2, r_3, r_4, r_5</math>. According to [[Vieta's formulas]], we have <math>d=r_1r_2r_3r_4 + r_1r_2r_3r_5 + r_1r_2r_4r_5 + r_1r_3r_4r_5 + r_2r_3r_4r_5</math>. The first four terms contain <math>r_1=0</math> and are therefore zero, thus <math>d=r_2r_3r_4r_5</math>. This is a product of four non-zero numbers, therefore <math>d</math> must be non-zero <math>\Longrightarrow \mathrm{(D)}</math>.
  
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 ===
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.
  
 
== See Also ==
 
== See Also ==
*[[2003 AMC 12A Problems]]
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{{AMC12 box|year=2003|ab=A|num-b=20|num-a=22}}
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[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Revision as of 00:58, 19 October 2020

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

Let the roots be $r_1=0, r_2, r_3, r_4, r_5$. According to Vieta's formulas, we have $d=r_1r_2r_3r_4 + r_1r_2r_3r_5 + r_1r_2r_4r_5 + r_1r_3r_4r_5 + r_2r_3r_4r_5$. The first four terms contain $r_1=0$ and are therefore zero, thus $d=r_2r_3r_4r_5$. This is a product of four non-zero numbers, therefore $d$ must be non-zero $\Longrightarrow \mathrm{(D)}$.

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 (ProblemsAnswer KeyResources)
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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