AMC12 2001 A

AMC12 2001 A · Q19

AMC12 2001 A · Q19. It mainly tests Quadratic equations, Vieta / quadratic relationships (basic).

The polynomial $p(x) = x^3+ax^2+bx+c$ has the property that the average of its zeros, the product of its zeros, and the sum of its coefficients are all equal. The $y$-intercept of the graph of $y=p(x)$ is 2. What is $b$?

多项式 $p(x) = x^3+ax^2+bx+c$ 具有如下性质：它的零点的平均值、零点的乘积以及系数的和都相等。函数 $y=p(x)$ 的图像的 $y$ 轴截距为 2。求 $b$。

(A) -11 -11

(B) -10 -10

(D) 1 1

(E) 5 5

Answer

Correct choice: (A)

正确答案：（A）

Solution

We are given $c=2$. So the product of the roots is $-c = -2$ by Vieta's formulas. These also tell us that $\frac{-a}{3}$ is the average of the zeros, so $\frac{-a}3=-2 \implies a = 6$. We are also given that the sum of the coefficients is $-2$, so $1+6+b+2 = -2 \implies b=-11$. So the answer is $\fbox{A}$.

已知 $c=2$。由韦达定理，零点的乘积为 $-c=-2$。韦达定理还告诉我们零点的平均值为 $\frac{-a}{3}$，所以 $\frac{-a}3=-2 \implies a = 6$。又已知系数和为 $-2$，因此 $1+6+b+2 = -2 \implies b=-11$。答案为 $\fbox{A}$。

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