Difference between revisions of "2023 AMC 12B Problems/Problem 14"

Revision as of 19:44, 15 November 2023

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

$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 6 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 4$

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 $\boxed{\textbf{(A) 5}}$ .

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

2023 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

@@ Line 1: / Line 1: @@
+==Problem==
+For how many ordered pairs <math>(a,b)</math> of integers does the polynomial <math>x^3+ax^2+bx+6</math> have <math>3</math> distinct integer roots?
+<math>\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 6 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 4</math>
 ==Solution==