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?

$a$

$b$

$c$

$d$

$e$

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.