AMC12 2003 A

AMC12 2003 A · Q21

AMC12 2003 A · Q21. It mainly tests Polynomials, Functions basics.

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?

多项式 \[P(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + e\] 的图像有五个不同的 $x$ 截距，其中一个在 $(0,0)$。以下哪个系数不可能为零？

(A) a a

(B) b b

(D) d d

(E) e e

Answer

Correct choice: (D)

正确答案：（D）

Solution

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 $\Longrightarrow \mathrm{(D)}$. 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.

设其根为 $r_1=0, r_2, r_3, r_4, r_5$。根据韦达定理，有 $d=r_1r_2r_3r_4 + r_1r_2r_3r_5 + r_1r_2r_4r_5 + r_1r_3r_4r_5 + r_2r_3r_4r_5$。前四项都含有 $r_1=0$，因此都为 $0$，所以 $d=r_2r_3r_4r_5$。这是四个非零数的乘积，因此 $d$ 必须非零 $\Longrightarrow \mathrm{(D)}$。显然，由于 $(0,0)$ 是截距，$e$ 必须为 $0$。但如果 $d$ 为 $0$，则 $x^2$ 会整除该多项式，这意味着在 $0$ 处有重根，这不可能，因为五个根都互不相同。

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