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

(New page: == Problem 21 == The graph of the polynomial <math>P(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + e</math> has five distinct <math>x</math>-intercepts, one of which is at <math>(0,0)</math>. Whi...)
 
(→‎Solution)
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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>.
 
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>.
 
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>
 
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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== See Also ==
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*[[2003 AMC 12A Problems]]

Revision as of 22:20, 31 May 2008

Problem 21

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?

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

Solution

According to Vieta's Formula, 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. $\Rightarrow D$

See Also

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