Decide if a given polynomial is irreducible

Find all values of n \in \mathbb{Z} such that the following polynomial is irreducible: p(x) = 3x^2+2nx+12.


Suppose p(x) is reducible. Then it factors as a product of linear polynomials. By Gauss’ lemma, we can assume without loss of generality that these factors are in \mathbb{Z}[x]. Since the leading coefficient of p(x) is prime, we can say (again without loss of generality) that 3x^2 + 2nx + 12 = (3x+a)(x+b) = 3x^2 + (3b+a)x + ab. Comparing coefficients, we have ab = 12 and 3b+a = 2n. the first equation has 12 solutions in the integers: (\pm 1, \pm 12), (\pm 12, \pm 1), (\pm 2, \pm 6), (\pm 6, \pm 2), (\pm 3, \pm 4), and (\pm 4, \pm 3). Reducing the second equation mod 2, we see that a \equiv b mod 2. This reduces our set of candidate solutions to four. If (a,b) = (2,6), then n = 10 and indeed p(x) = (3x+2)(x+6). If (a,b) = (-2,-6), then n = -10 and indeed p(x) = (3x-2)(x-6). If (a,b) = (6,2), then n = 6 and indeed p(x) = (3x+6)(x+2). If (a,b) = (-6,-2), then n = -6, and indeed p(x) = (3x-6)(x-2).

For all other n, p(x) is irreducible.

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