Factor the following polynomial over $\mathbb{Z} [ X ] $ I need to factor the following polynomial  $ f \in \mathbb{Z} [ X ] $, and if I can't factor it out to give an argument why is irreductible.
$f=8x^4+18x^3-41x^2-68x+18$.  I tryeid using Horner's Method by finding the divizors of the free term (i.e. 18). I also tryied to find the discriminant of f using the following formulas $D(f) = \frac{(-1)^{\frac{n(n-1)}{2}}}{a_n}Res(f,f')$, where $Res(f,f')$ is the determinant of the Sylvester Matrix. If I could prove that the discriminant of $f$ is negative then $f$ would be irreductible.
 A: It's not difficult to show that $f$ does not have a rational root, i.e. a factor of degree $1$. (Use the rational root theorem, as mentioned in the comments.)
Thus to show that $f$ is irreducible over $\mathbb Z$, we just have to eliminate the possibility that $f=gh$, where $g$ and $h$ are polynomials over $\mathbb Z$ of degree $2$.
If we had such $g$ and $h$, then there could be at most $8$ values $n \in \mathbb Z$, such that $f(n)$ is prime. This is because we have $f(n) = g(n) \cdot h(n)$. Now if $f(n)$ is prime, this means that either $g(n)$ or $h(n)$ is $\pm 1$. Since $g\pm1$ and $h\pm 1$ are each polynomials of degree $2$, they can have at most $2$ roots, thus both $g$ and $h$ can attain each of the values $\pm 1$ at most twice. If we combine the possibilities for positive and negative $1$, each $g$ and $h$ can attain a unit value in $\mathbb Z$ for at most $4$ values, so in total there can be at most $8$ values $n$ such that $f(n)$ is prime. (So far this argument works for any degree $4$ integer polynomial with no rational roots.)
So to complete the proof of irreducibility, we just have to find $9$ values $n \in \mathbb Z$ such that $f(n)$ is prime.
$f(5)=5903$, $f(-5)=2083$, $f(-7)=11519$, $f(35)=12724163$, $f(-35)=11185423$,  $f(-37)=14027939$, $f(-97)=691426979$, $f(103)=919634179$, $f(-115)=1371294863$ are all prime values.
Thus $f$ is irreducible over $\mathbb Z$ (and by Gauss's lemma also over $\mathbb Q$).
