Prove that f(x) = x^4 - 2*x^2 + 8*x + 1 is irreducible in Q

Prove that f(x) = x^4 - 2*x^2 + 8*x + 1 is irreducible in Q.

We show that f is irreducible over Z[x].
(By Gauss' Lemma, f being irreducible over Z[x] ==> f is irreducible over Q[x].

By the Rational Root Theorem, x = -1, 1 are the only possible rational roots of f.
Since neither of them are actually roots of f, we conclude that f has no linear factors.
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Since f is of degree 4, there remains the possibility that f(x) is the product of two
quadratic factors.

Hence, f(x) = (x^2 + ax + b)(x^2 + cx + d) for some a,b,c,d in Z.
So, we obtain
x^4 - 2x^2 + 8x + 1 = x^4 + (a+c) x^3 + (b + ac + d)x^2 + (bc + ad) x + bc.

Equating like coefficients,
a + c = 0 ==> c = -a
b + ac + d = -2 ==> b - a^2 + d = -2
bc + ad = 8 ==> a(d - b) = 8
bc = 1 ==> b = c = ± 1.

If b = c = 1, then a = -1.
==> 1 - (-1)^2 + d = -2 ==> d = -2.
However, then a(d - b) ≠ -2, a contradiction.

If b = c = -1, then a = 1.
==> -1 - 1^2 + d = -2 ==> d = 0.
However, then a(d - b) ≠ -2, a contradiction.

Hence, f can't be written as a product of quadratic factors as well.
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I hope this helps!

By the Rational Root Theorem, possible zeroes of f are ±1. Neither is a zero, so f has no rational zeroes and is thus irreducible over ℚ.