Asked by crystal
Which polynomial is prime?
x3 + 3x2 – 2x – 6
x3 – 2x2 + 3x – 6
4x4 + 4x3 – 2x – 2
2x4 + x3 – x + 2
All Answers 1
Answered by
GPT-5 mini
AI
Prime (irreducible over Q): 2x^4 + x^3 − x + 2.
Reasoning:
- x^3 + 3x^2 − 2x − 6 = (x+3)(x^2 − 2).
- x^3 − 2x^2 + 3x − 6 = (x−2)(x^2 + 3).
- 4x^4 + 4x^3 − 2x − 2 = (x+1)(4x^3 − 2) = 2(x+1)(2x^3 − 1).
For 2x^4 + x^3 − x + 2: there is no rational linear root (±1, ±2, ±1/2 all fail). If it factored nontrivially over Q it would factor as two integer quadratics (Gauss's lemma). Writing (ax^2+bx+c)(dx^2+ex+f) with ad = 2 and cf = 2 and checking the finite possibilities for (a,d) and (c,f) leads to no integer solution for the remaining coefficient equations. Hence it has no linear or quadratic factor over Q and is irreducible.
Reasoning:
- x^3 + 3x^2 − 2x − 6 = (x+3)(x^2 − 2).
- x^3 − 2x^2 + 3x − 6 = (x−2)(x^2 + 3).
- 4x^4 + 4x^3 − 2x − 2 = (x+1)(4x^3 − 2) = 2(x+1)(2x^3 − 1).
For 2x^4 + x^3 − x + 2: there is no rational linear root (±1, ±2, ±1/2 all fail). If it factored nontrivially over Q it would factor as two integer quadratics (Gauss's lemma). Writing (ax^2+bx+c)(dx^2+ex+f) with ad = 2 and cf = 2 and checking the finite possibilities for (a,d) and (c,f) leads to no integer solution for the remaining coefficient equations. Hence it has no linear or quadratic factor over Q and is irreducible.
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.