2003 AMC 12A 第 21 题

Question

多项式

\begin{aligned} &P(x) = x^5 + ax^4 + bx^3 \\ &\quad {}+ cx^2 + dx + e \end{aligned}

的图像有五个不同的

x

截距，其中一个是

(0, 0)

。下列哪个系数不可能为零？

The graph of the polynomial

\begin{aligned} &P(x) = x^5 + ax^4 + bx^3 \\ &\quad {}+ cx^2 + dx + e \end{aligned}

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

P(0)=0

迫使

e=0

。

P(0)=0

forces

e=0

提出因子

x

；剩下四次多项式的常数项是四个非零根的乘积。

Factor out

x;

the constant term of the remaining quartic is the product of the four nonzero roots

因为

(0,0)

是截距，

P(0)=e=0

，所以

P(x)

=x\left(x^4+ax^3+bx^2+cx+d\right)

。

其余四个截距都是非零且互不相同的根，而

d

等于这四个非零根的乘积，因此不可能为零。

通过适当选择根，

a,b,c

都可能为零，但

d\neq0

。

所以正确答案是 D。

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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