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

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== Solution 2 ==
 
== Solution 2 ==
  
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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.
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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.
  
 
== See Also ==
 
== See Also ==
 
*[[2003 AMC 12A Problems]]
 
*[[2003 AMC 12A Problems]]

Revision as of 20:51, 12 June 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. $\Rightarrow D$

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

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