2003 AMC 12A Problems/Problem 21: Difference between revisions

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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 Viète's formulae, 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.

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

@@ Line 1: / Line 1: @@
-== Problem 21 ==
+== Problem ==
-The graph of the polynomial
+The graph of the [[polynomial]]
-<math>P(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + e</math>
+<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?
+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>
+<math>\text{(A)}\ a \qquad \text{(B)}\ b \qquad \text{(C)}\ c \qquad \text{(D)}\ d \qquad \text{(E)}\ e</math>
 == Solution ==
+=== Solution 1 ===
+Let the roots be <math>r_1=0, r_2, r_3, r_4, r_5</math>. According to [[Vieta's formulas | Viète's formulae]], 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>.
+=== 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>
+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.
-== Solution 2 ==
+== See Also ==
+{{AMC12 box|year=2003|ab=A|num-b=20|num-a=22}}
-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.
+[[Category:Intermediate Algebra Problems]]
+{{MAA Notice}}
-== See Also ==
-*[[2003 AMC 12A Problems]]

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