2003 AMC 12A Problem 21 — Solution | LIVE by Po-Shen Loh

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2003 AMC 12A Problem 21

Below is the professionally curated solution for Problem 21 of the 2003 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 12A solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:Vieta’s Formulas polynomial

Difficulty rating: 1990

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?

$a$

$b$

$c$

$d$

$e$

Solution:

Since $(0,0)$ is an intercept, $P(0)=e=0,$ so $P(x)=x\left(x^4+ax^3+bx^2+cx+d\right).$

The four remaining intercepts are nonzero and distinct, and $d$ equals their product, which is therefore nonzero.

Any of $a,b,c$ can be zero for suitable choices of those roots, but $d\neq0.$

Thus, the correct answer is D.

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