Irreducibility of $f(x)=x^4+3x^3-9x^2+7x+27$ Question at hand is:

Is $x^4+3x^3-9x^2+7x+27$ irreducible in $\Bbb Q$ and/or $\Bbb Z$.

This is for an exam, reasoning is trivial, but no calculators in hand. Clearly, if there is a rational root, they are integers by Rational Root theorem and since $f$ is monic.
I am aware of 


*

*Rational root theorem, which narrows down the options to $\pm1,\pm3,\pm9,\pm27$, and clearly, no roots.

*Eisenstein's Irreducibility Criteria, not helping here, thanks to $x$'s coefficient $7$

*Cohn's Irreducibility test: $12197$ is a prime, too large a number to prove that its a prime by hand.

*Descartes Rule of signs: at most 2 (or 0) positive/negative roots. Close enough. 
None of which are helping me in any way since I can't use a calculator. 
These are the solutions I tried:


*

*Alpha says all roots are complex. Made me search if there's some way to determine if all roots are complex, reaching nowhere.

*Check if there are any easy prime generation functions like Euler's, and if lucky 12197 falls in that list, the best I got is Euler's, $n^2+n+41, 1\le n<40$, and biggest such is $1601$, not helping.
Are there any better ways to determine if this polynomial is irreducible over $\Bbb Q$, without using calculators?
 A: Over $\mathbb{F}_3$ your polynomial splits as $x(x+1)^3$, no luck. 
Over $\mathbb{F}_5$ your polynomial splits as $(x+1)(x^3+x^2+2)$. That is enough to state that your polynomial is irreducible over $\mathbb{Q}$, since we know in advance that it has no rational root.
You may also notice that $p(x)=x^4 + 3 x^3 - 9 x^2 + 7 x + 27$ takes prime values for any $x\in\{-20,-17,-14,-5,-2,1,4,10,13\}$, but no reducible polynomial over $\mathbb{Q}$ with degree $4$ can take nine prime values at nine different points.
A: Echoing the answer by Prometheus note that modulo $2$ the reduced polynomial is $\Phi_5$, which is irreducible modulo $2$, since $2$ has order $4$ in $\Bbb Z_5^\times$: $2^2=4,2^3=8,2^4=16=1$. In general the $n$-th cyclotomic polynomial $\Phi_n$ is irreducible in $\Bbb F_q[X]$ iff $q$ has order $\varphi(n)$ in $\Bbb Z_n^\times$. 
A: If you take the polynomial modulo 2 you get $x^4+x^3+x^2+x+1$. It doesn't have a root, so if it is decomposable, it is a product of two irreducible polynomials of degree 2. The only polynomial of degree 2 which doesn't have a root is $x^2+x+1$ and its square is $x^4+x^2+1$. It follows that your polynomial is irreducible modulo $2$ and therefore irreducible in over $\mathbb{Z}$.
A: Irreducibility over $\mathbb Q$ is the same as over $\mathbb Z$ (Gauss lemma). There is no root (real or complex) not exceeding $1$ in absolute value ($27>1+3+9+7$). Thus, if there is a factorization into two polynomials, the free terms must be $\pm 3$ and $\pm 9$ (otherwise, by Vieta, the polynomial with the free term $\pm 1$ has a root with absolute value $\le 1$). However then the coefficient at $x$ must be divisible by $3$ and it is not. 
A: Theorem. (Gauss). If $p(x)\in Z[x]$ is reducible over $Q$ then $p(x)$ is reducible over Z. And to show that $p(x)=x^4+3x^3-9x^2+7x+27$ is not the product of two quadratics in $Z[x]$, it suffices to show that if $p(x)=(x^2+A x+B)(x^2+C x+D)$ with four cases: $(B,D)\in \{(1,27),(-1,-27),(3,9),(-3,-9)\}$,.... then  $A+B=3$ and $B+D+AC=-9$ and $ A D+B C=7$, which cannot be solved in integers $A,C$.  
