# 2003 AMC 12A Problems/Problem 21

## 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.

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