金声玉亮2.0模型评测报告

问题

There is a single choice question about abstract algebra. Answer the question by replying A, B, C or D.
Question: Determine whether the polynomial in Z[x] satisfies an Eisenstein criterion for irreducibility over Q. 8x^3 + 6x^2 - 9x + 24
A. Yes, with p=2.
B. Yes, with p=3.
C. Yes, with p=5.
D. No.
Answer:

答案

模型	模型答案	正确答案
金声玉亮2.0	To apply the Eisenstein criterion for irreducibility, we need to find a prime number $p$ such that: 1. $p$ does not divide the leading coefficient of the polynomial (in this case, 8). 2. $p$ divides all the other coefficients of the polynomial (in this case, 6, -9, and 24). 3. $p^2$ does not divide the constant term of the polynomial (in this case, 24). Let's check the options: A. $p=2$: * 2 does not divide 8, so condition 1 is satisfied. * 2 divides 6, -9, and 24, so condition 2 is satisfied. * 2^2 = 4 does not divide 24, so condition 3 is satisfied. Therefore, the polynomial satisfies the Eisenstein criterion for irreducibility over Q with $p=2$. The correct answer is: A. Yes, with p=2.	B
deepseekr1	The polynomial \(8x^3 + 6x^2 - 9x + 24\) is analyzed using Eisenstein's criterion. For each prime option: - Option A (p=2): Fails because 2 does not divide \(-9\) (coefficient of \(x\)), violating condition 1. - Option B (p=3): - All non-leading coefficients (6, -9, 24) are divisible by 3. - The leading coefficient (8) is not divisible by 3. - \(3^2 = 9\) does not divide the constant term (24). All three conditions are satisfied, so Eisenstein applies with \(p=3\). - Option C (p=5): Fails because 5 does not divide 6 (coefficient of \(x^2\)), violating condition 1. - Option D (No): Incorrect since \(p=3\) works. Answer: B	B
gpto1	B	B
GPT-4o	D. No.	B
llama3.1-70B-Instruct	A	B
llama3.1-405B-Instruct	To apply the Eisenstein criterion for irreducibility, we need to find a prime number $p$ such that the following conditions are satisfied: 1. $p$ does not divide the leading coefficient of the polynomial. 2. $p$ divides all the other coefficients of the polynomial. 3. $p^2$ does not divide the constant term of the polynomial. Let's examine the polynomial $8x^3 + 6x^2 - 9x + 24$: 1. The leading coefficient is 8, so we need to find a prime number that does not divide 8. The prime numbers that do not divide 8 are 3, 5, 7, 11, ... 2. The other coefficients are 6, -9, and 24. We need to find a prime number that divides all of these coefficients. The prime numbers that divide 6 are 2 and 3. The prime numbers that divide -9 are 3. The prime numbers that divide 24 are 2 and 3. Therefore, the only prime number that divides all the other coefficients is 3. 3. The constant term is 24, and we need to find a prime number such that $p^2$ does not divide 24. Since $3^2 = 9$ does not divide 24, the prime number 3 satisfies this condition. Therefore, the polynomial $8x^3 + 6x^2 - 9x + 24$ satisfies the Eisenstein criterion for irreducibility over $\mathbb{Q}$ with $p=3$. The correct answer is B.	B
DeepSeek-Coder-V2	B	B
Qwen2-72B-Instruct	B. Yes, with p=3.	B
baichuan4	B. Yes, with p=3.	B
doubao	D. No.	B
minimax	To determine whether the polynomial \(8x^3 + 6x^2 - 9x + 24\) satisfies an Eisenstein criterion for irreducibility over \(\mathbb{Q}\), we need to find a prime number \(p\) such that the following conditions are met: 1. \(p\) divides all coefficients of the polynomial except the leading coefficient. 2. \(p^2\) does not divide the constant term. 3. \(p\) does not divide the leading coefficient. Let's check the coefficients of the polynomial \(8x^3 + 6x^2 - 9x + 24\): - The leading coefficient is 8. - The constant term is 24. - The other coefficients are 6, -9. We can see that 2 divides all coefficients except the leading coefficient, and \(2^2 = 4\) does not divide the constant term 24. However, 2 does divide the leading coefficient 8, so the Eisenstein criterion with \(p=2\) does not apply. Next, let's check with \(p=3\): - 3 divides 6 and -9, but not 8 (the leading coefficient). - \(3^2 = 9\) does not divide 24 (the constant term). Since 3 divides all coefficients except the leading coefficient and \(3^2\) does not divide the constant term, and 3 does not divide the leading coefficient, the polynomial satisfies the Eisenstein criterion with \(p=3\). Therefore, the correct answer is: B. Yes, with p=3.	B
kimi	To determine whether the polynomial \( P(x) = 8x^3 + 6x^2 - 9x + 24 \) satisfies the Eisenstein criterion for irreducibility over the rational numbers \( \mathbb{Q} \), we need to check if there exists a prime number \( p \) such that: 1. \( p \) divides the coefficient of every term except for the leading coefficient. 2. \( p \) does not divide the leading coefficient. 3. \( p^2 \) does not divide the constant term. Let's check each option: A. \( p = 2 \): - \( 2 \) divides \( 6 \) (coefficient of \( x^2 \)), \( -9 \) (coefficient of \( x \)), and \( 24 \) (constant term). - \( 2 \) does not divide \( 8 \) (leading coefficient). - \( 2^2 = 4 \) does not divide \( 24 \) (constant term). The condition is satisfied for \( p = 2 \), so the polynomial satisfies the Eisenstein criterion for irreducibility over \( \mathbb{Q} \). Answer: A. Yes, with p=2.	B
wenxinyiyan	B Explanation: To apply Eisenstein's criterion, we need to find a prime number p that divides all coefficients of the polynomial except the highest degree term, and p^2 doesn't divide the constant term. Let's check each option: A. For p=2, it divides 8, 6, and 24, but it also divides the coefficient of x^3 (which is 8). So, it doesn't satisfy the Eisenstein criterion. B. For p=3, it divides 6 and 24, but it doesn't divide 8 (coefficient of x^3) and 3^2 = 9 doesn't divide 24 (the constant term). So, it satisfies the Eisenstein criterion. C. For p=5, it doesn't divide all coefficients except the leading one. So, it doesn't satisfy the criterion. D. Since we found a prime (p=3) that satisfies the Eisenstein criterion, this option is not correct. Therefore, the correct answer is B.	B
zhipu	To determine whether the polynomial \( 8x^3 + 6x^2 - 9x + 24 \) satisfies the Eisenstein criterion for irreducibility over \( \mathbb{Q} \), we need to check the conditions of the criterion for some prime \( p \). The Eisenstein criterion states that a polynomial \( a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \) with integer coefficients is irreducible over \( \mathbb{Q} \) if there exists a prime \( p \) such that: 1. \( p \) divides \( a_i \) for all \( i \neq n \), 2. \( p \) does not divide \( a_n \), 3. \( p^2 \) does not divide \( a_0 \). Let's analyze the polynomial \( 8x^3 + 6x^2 - 9x + 24 \) for different primes \( p \). ### Checking \( p = 2 \): - \( a_3 = 8 \): \( 2 \) divides \( 8 \). - \( a_2 = 6 \): \( 2 \) divides \( 6 \). - \( a_1 = -9 \): \( 2 \) does not divide \( -9 \). Since \( 2 \) does not divide \( a_1 \), the Eisenstein criterion fails for \( p = 2 \). ### Checking \( p = 3 \): - \( a_3 = 8 \): \( 3 \) does not divide \( 8 \). Since \( 3 \) does not divide \( a_3 \), the Eisenstein criterion fails for \( p = 3 \). ### Checking \( p = 5 \): - \( a_3 = 8 \): \( 5 \) does not divide \( 8 \). Since \( 5 \) does not divide \( a_3 \), the Eisenstein criterion fails for \( p = 5 \). ### Conclusion: None of the primes \( 2, 3, \) or \( 5 \) satisfy the Eisenstein criterion for the polynomial \( 8x^3 + 6x^2 - 9x + 24 \). Therefore, the polynomial does not satisfy the Eisenstein criterion for any prime \( p \). The correct answer is: D. No.	B
gpto1	B	B
deepseekv3	To determine whether the polynomial \( f(x) = 8x^3 + 6x^2 - 9x + 24 \) satisfies the Eisenstein criterion for irreducibility over \( \mathbb{Q} \), we need to check if there exists a prime number \( p \) such that: 1. \( p \) divides all the coefficients except the leading coefficient. 2. \( p^2 \) does not divide the constant term. Let's analyze the coefficients with respect to the given primes: - For \( p = 2 \): - Coefficients: \( 8, 6, -9, 24 \). - \( 2 \) divides \( 8, 6, 24 \), but it does not divide the coefficient \( -9 \). Thus, \( p = 2 \) does not satisfy the first condition. - For \( p = 3 \): - Coefficients: \( 8, 6, -9, 24 \). - \( 3 \) divides \( 6, -9, 24 \), but it does not divide the leading coefficient \( 8 \). Thus, \( p = 3 \) does not satisfy the first condition. - For \( p = 5 \): - Coefficients: \( 8, 6, -9, 24 \). - \( 5 \) does not divide \( 8, 6, -9, 24 \). Thus, \( p = 5 \) does not satisfy the first condition. Since there is no prime \( p \) that satisfies the Eisenstein criterion for this polynomial, the correct answer is: D. No.	B

模型

模型答案

正确答案

金声玉亮2.0

To apply the Eisenstein criterion for irreducibility, we need to find a prime number $p$ such that:

1. $p$ does not divide the leading coefficient of the polynomial (in this case, 8).
2. $p$ divides all the other coefficients of the polynomial (in this case, 6, -9, and 24).
3. $p^2$ does not divide the constant term of the polynomial (in this case, 24).

Let's check the options:

A. $p=2$:

* 2 does not divide 8, so condition 1 is satisfied.
* 2 divides 6, -9, and 24, so condition 2 is satisfied.
* 2^2 = 4 does not divide 24, so condition 3 is satisfied.

Therefore, the polynomial satisfies the Eisenstein criterion for irreducibility over Q with $p=2$.

The correct answer is:

A. Yes, with p=2.

deepseekr1

The polynomial $8x^3 + 6x^2 - 9x + 24$ is analyzed using Eisenstein's criterion. For each prime option:

- **Option A (p=2)**: Fails because 2 does not divide $-9$ (coefficient of $x$), violating condition 1.
- **Option B (p=3)**:
- All non-leading coefficients (6, -9, 24) are divisible by 3.
- The leading coefficient (8) is not divisible by 3.
- $3^2 = 9$ does not divide the constant term (24).
All three conditions are satisfied, so Eisenstein applies with $p=3$.
- **Option C (p=5)**: Fails because 5 does not divide 6 (coefficient of $x^2$), violating condition 1.
- **Option D (No)**: Incorrect since $p=3$ works.

**Answer: B**

gpto1

GPT-4o

D. No.

llama3.1-70B-Instruct

llama3.1-405B-Instruct

To apply the Eisenstein criterion for irreducibility, we need to find a prime number $p$ such that the following conditions are satisfied:

1. $p$ does not divide the leading coefficient of the polynomial.
2. $p$ divides all the other coefficients of the polynomial.
3. $p^2$ does not divide the constant term of the polynomial.

Let's examine the polynomial $8x^3 + 6x^2 - 9x + 24$:

1. The leading coefficient is 8, so we need to find a prime number that does not divide 8. The prime numbers that do not divide 8 are 3, 5, 7, 11, ...
2. The other coefficients are 6, -9, and 24. We need to find a prime number that divides all of these coefficients. The prime numbers that divide 6 are 2 and 3. The prime numbers that divide -9 are 3. The prime numbers that divide 24 are 2 and 3. Therefore, the only prime number that divides all the other coefficients is 3.
3. The constant term is 24, and we need to find a prime number such that $p^2$ does not divide 24. Since $3^2 = 9$ does not divide 24, the prime number 3 satisfies this condition.

Therefore, the polynomial $8x^3 + 6x^2 - 9x + 24$ satisfies the Eisenstein criterion for irreducibility over $\mathbb{Q}$ with $p=3$.

The correct answer is B.

DeepSeek-Coder-V2

Qwen2-72B-Instruct

B. Yes, with p=3.

baichuan4

B. Yes, with p=3.

doubao

D. No.

minimax

To determine whether the polynomial $8x^3 + 6x^2 - 9x + 24$ satisfies an Eisenstein criterion for irreducibility over $\mathbb{Q}$, we need to find a prime number $p$ such that the following conditions are met:

1. $p$ divides all coefficients of the polynomial except the leading coefficient.
2. $p^2$ does not divide the constant term.
3. $p$ does not divide the leading coefficient.

Let's check the coefficients of the polynomial $8x^3 + 6x^2 - 9x + 24$:

- The leading coefficient is 8.
- The constant term is 24.
- The other coefficients are 6, -9.

We can see that 2 divides all coefficients except the leading coefficient, and $2^2 = 4$ does not divide the constant term 24. However, 2 does divide the leading coefficient 8, so the Eisenstein criterion with $p=2$ does not apply.

Next, let's check with $p=3$:

- 3 divides 6 and -9, but not 8 (the leading coefficient).
- $3^2 = 9$ does not divide 24 (the constant term).

Since 3 divides all coefficients except the leading coefficient and $3^2$ does not divide the constant term, and 3 does not divide the leading coefficient, the polynomial satisfies the Eisenstein criterion with $p=3$.

Therefore, the correct answer is:

B. Yes, with p=3.

kimi

To determine whether the polynomial $ P(x) = 8x^3 + 6x^2 - 9x + 24 $ satisfies the Eisenstein criterion for irreducibility over the rational numbers $ \mathbb{Q} $, we need to check if there exists a prime number $ p $ such that:

1. $ p $ divides the coefficient of every term except for the leading coefficient.
2. $ p $ does not divide the leading coefficient.
3. $ p^2 $ does not divide the constant term.

Let's check each option:

A. $ p = 2 $:
- $ 2 $ divides $ 6 $ (coefficient of $ x^2 $), $ -9 $ (coefficient of $ x $), and $ 24 $ (constant term).
- $ 2 $ does not divide $ 8 $ (leading coefficient).
- $ 2^2 = 4 $ does not divide $ 24 $ (constant term).

The condition is satisfied for $ p = 2 $, so the polynomial satisfies the Eisenstein criterion for irreducibility over $ \mathbb{Q} $.

Answer: A. Yes, with p=2.

wenxinyiyan

B

Explanation:
To apply Eisenstein's criterion, we need to find a prime number p that divides all coefficients of the polynomial except the highest degree term, and p^2 doesn't divide the constant term.

Let's check each option:

A. For p=2, it divides 8, 6, and 24, but it also divides the coefficient of x^3 (which is 8). So, it doesn't satisfy the Eisenstein criterion.

B. For p=3, it divides 6 and 24, but it doesn't divide 8 (coefficient of x^3) and 3^2 = 9 doesn't divide 24 (the constant term). So, it satisfies the Eisenstein criterion.

C. For p=5, it doesn't divide all coefficients except the leading one. So, it doesn't satisfy the criterion.

D. Since we found a prime (p=3) that satisfies the Eisenstein criterion, this option is not correct.

Therefore, the correct answer is B.

zhipu

To determine whether the polynomial $ 8x^3 + 6x^2 - 9x + 24 $ satisfies the Eisenstein criterion for irreducibility over $ \mathbb{Q} $, we need to check the conditions of the criterion for some prime $ p $.

The Eisenstein criterion states that a polynomial $ a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 $ with integer coefficients is irreducible over $ \mathbb{Q} $ if there exists a prime $ p $ such that:
1. $ p $ divides $ a_i $ for all $ i \neq n $,
2. $ p $ does not divide $ a_n $,
3. $ p^2 $ does not divide $ a_0 $.

Let's analyze the polynomial $ 8x^3 + 6x^2 - 9x + 24 $ for different primes $ p $.

### Checking $ p = 2 $:
- $ a_3 = 8 $: $ 2 $ divides $ 8 $.
- $ a_2 = 6 $: $ 2 $ divides $ 6 $.
- $ a_1 = -9 $: $ 2 $ does not divide $ -9 $.

Since $ 2 $ does not divide $ a_1 $, the Eisenstein criterion fails for $ p = 2 $.

### Checking $ p = 3 $:
- $ a_3 = 8 $: $ 3 $ does not divide $ 8 $.

Since $ 3 $ does not divide $ a_3 $, the Eisenstein criterion fails for $ p = 3 $.

### Checking $ p = 5 $:
- $ a_3 = 8 $: $ 5 $ does not divide $ 8 $.

Since $ 5 $ does not divide $ a_3 $, the Eisenstein criterion fails for $ p = 5 $.

### Conclusion:
None of the primes $ 2, 3, $ or $ 5 $ satisfy the Eisenstein criterion for the polynomial $ 8x^3 + 6x^2 - 9x + 24 $. Therefore, the polynomial does not satisfy the Eisenstein criterion for any prime $ p $.

The correct answer is:
D. No.

gpto1

deepseekv3

To determine whether the polynomial $ f(x) = 8x^3 + 6x^2 - 9x + 24 $ satisfies the Eisenstein criterion for irreducibility over $ \mathbb{Q} $, we need to check if there exists a prime number $ p $ such that:

1. $ p $ divides all the coefficients except the leading coefficient.
2. $ p^2 $ does not divide the constant term.

Let's analyze the coefficients with respect to the given primes:

- **For $ p = 2 $:**
- Coefficients: $ 8, 6, -9, 24 $.
- $ 2 $ divides $ 8, 6, 24 $, but it does not divide the coefficient $ -9 $. Thus, $ p = 2 $ does not satisfy the first condition.

- **For $ p = 3 $:**
- Coefficients: $ 8, 6, -9, 24 $.
- $ 3 $ divides $ 6, -9, 24 $, but it does not divide the leading coefficient $ 8 $. Thus, $ p = 3 $ does not satisfy the first condition.

- **For $ p = 5 $:**
- Coefficients: $ 8, 6, -9, 24 $.
- $ 5 $ does not divide $ 8, 6, -9, 24 $. Thus, $ p = 5 $ does not satisfy the first condition.

Since there is no prime $ p $ that satisfies the Eisenstein criterion for this polynomial, the correct answer is:

**D. No.**