2023 AMC 12B Problems/Problem 14: Difference between revisions
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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== | ==Solution== | ||
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Putting all cases together, the total number of solutions is | Putting all cases together, the total number of solutions is | ||
\boxed{\textbf{(A) 5}}. | <math>\boxed{\textbf{(A) 5}}</math>. | ||
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~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 | |
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