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

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==Problem==
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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?
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<math>\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 6 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 4</math>
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==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
 
<math>\boxed{\textbf{(A) 5}}</math>.
 
<math>\boxed{\textbf{(A) 5}}</math>.
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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==Video Solution 1 by OmegaLearn==
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https://youtu.be/sgVkR0AOGhE
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==Video Solution==
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https://youtu.be/IS6BDxGYTsE
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Latest revision as of 01:53, 20 November 2023

Problem

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

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

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

~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

2023 AMC 12B (ProblemsAnswer KeyResources)
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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