Difference between revisions of "2023 AMC 12B Problems/Problem 14"
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| + | ==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== | ||
| + | |||
| + | Denote three roots as <math>r_1 < r_2 < r_3</math>. | ||
| + | Following from Vieta's formula, <math>r_1r_2r_3 = -6</math>. | ||
| + | |||
| + | Case 1: All roots are negative. | ||
| + | |||
| + | We have the following solution: <math>\left( -3, -2, -1 \right)</math>. | ||
| + | |||
| + | Case 2: One root is negative and two roots are positive. | ||
| + | |||
| + | We have the following solutions: <math>\left( -3, 1, 2 \right)</math>, <math>\left( -2, 1, 3 \right)</math>, <math>\left( -1, 2, 3 \right)</math>, <math>\left( -1, 1, 6 \right)</math>. | ||
| + | |||
| + | Putting all cases together, the total number of solutions is | ||
| + | <math>\boxed{\textbf{(A) 5}}</math>. | ||
| + | |||
| + | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
| + | |||
| + | ==Video Solution 1 by OmegaLearn== | ||
| + | https://youtu.be/sgVkR0AOGhE | ||
| + | |||
| + | ==Video Solution== | ||
| + | |||
| + | https://youtu.be/IS6BDxGYTsE | ||
| + | |||
| + | |||
| + | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC12 box|year=2023|ab=B|num-b=13|num-a=15}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 01:53, 20 November 2023
Problem
For how many ordered pairs
of integers does the polynomial
have
distinct integer roots?
Solution
Denote three roots as
.
Following from Vieta's formula,
.
Case 1: All roots are negative.
We have the following solution:
.
Case 2: One root is negative and two roots are positive.
We have the following solutions:
,
,
,
.
Putting all cases together, the total number of solutions is
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution 1 by OmegaLearn
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
| 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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.