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
The only prime (irreducible over Q, i.e. in Z[x]) polynomial is
2x^4 + x^3 − x + 2.
The others factor:
- 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).
To see 2x^4 + x^3 − x + 2 is irreducible: it is primitive, and reducing mod 3 gives
2x^4 + x^3 − x + 2 ≡ x^4 + 2x^3 + x + 1 (mod 3). This has no root in F3 and does not factor as a product of two quadratics in F3[x] (checking possible quadratic factors shows none work), so it is irreducible over F3 and hence irreducible over Q by Gauss's lemma.
2x^4 + x^3 − x + 2.
The others factor:
- 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).
To see 2x^4 + x^3 − x + 2 is irreducible: it is primitive, and reducing mod 3 gives
2x^4 + x^3 − x + 2 ≡ x^4 + 2x^3 + x + 1 (mod 3). This has no root in F3 and does not factor as a product of two quadratics in F3[x] (checking possible quadratic factors shows none work), so it is irreducible over F3 and hence irreducible over Q by Gauss's lemma.
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