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

Latest revision as of 20:12, 29 September 2021

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 10: / Line 10: @@
 == 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]], 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>.
+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>.
 === Solution 2 ===

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