As mentioned, I am trying to show that $x^6 + 69x^5 − 511x + 363$ is irreducible over $\mathbb Z$. To see that it has no roots and no cubic factors, I send the polynomial to $\mathbb F_7$ and $\mathbb F_2$ respectively. The thing I have not figured out is how to show it has no quadratic factors. Is there any way to do this other than brute force?
$x^6 + 69x^5 − 511x + 363$ is irreducible over $\mathbb Z$
abstract-algebrairreducible-polynomialspolynomials
Best Answer
Adding $\mathbb F_3$ and $\mathbb F_{11}$ to the mix, the constant term $c$ of a quadratic factor of $p(x)$ must satisfy:
$p(x) = (x + 1)^2 (x^4 + x^3 + x^2 + x + 1)$ over $\mathbb F_2$ implies $c \equiv 1 \pmod{2}$
$p(x) = x (x - 1) (x^4 + x^3 + x^2 + x + 1)$ over $\mathbb F_3$ implies $c \equiv 0 \pmod{3}$
$p(x) = (x^2 - x + 4) (x^4 + 3 x^2 + 3 x + 5)$ over $\mathbb F_7$ implies $c \equiv 4 \pmod{7}$
$p(x) = x (x + 2) (x^4 + x^3 + 9 x^2 + 4 x + 3)$ over $\mathbb F_{11}$ implies $c \equiv 0 \pmod{11}$
The general solution to the above is $c \equiv 165 \pmod{462}$, but there is no such $c$ that divides $363$, the constant term of $p(x)$, so no quadratic factor exists.