AMC12 2023 B

AMC12 2023 B · Q14

AMC12 2023 B · Q14. It mainly tests Vieta / quadratic relationships (basic), Divisibility & factors.

For how many ordered pairs $(a,b)$ of integers does the polynomial $x^3+ax^2+bx+6$ have $3$ distinct integer roots?

多项式$x^3+ax^2+bx+6$具有$3$个不同的整数根，有多少个整数有序对$(a,b)$？

(A) 5 5

(B) 6 6

(D) 7 7

(E) 4 4

Answer

Correct choice: (A)

正确答案：（A）

Solution

Denote three roots as $r_1 < r_2 < r_3$. Following from Vieta's formula, $r_1r_2r_3 = -6$. Case 1: All roots are negative. We have the following solution: $\left( -3, -2, -1 \right)$. Case 2: One root is negative and two roots are positive. We have the following solutions: $\left( -3, 1, 2 \right)$, $\left( -2, 1, 3 \right)$, $\left( -1, 2, 3 \right)$, $\left( -1, 1, 6 \right)$. Putting all cases together, the total number of solutions is $\boxed{\textbf{(A) 5}}$.

设三个根为$r_1 < r_2 < r_3$。由Vieta公式，$r_1r_2r_3 = -6$。情况1：所有根均为负数。有解：$\left( -3, -2, -1 \right)$。情况2：一个根为负，两个根为正。有解：$\left( -3, 1, 2 \right)$，$\left( -2, 1, 3 \right)$，$\left( -1, 2, 3 \right)$，$\left( -1, 1, 6 \right)$。总共有$\boxed{\textbf{(A) 5}}$个解。

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