Showing that a polynomial is irreducible Assume that $p$ is a prime number. I'm to show that the polynomial
$f(x) = x^4 + x + p$ is irreducible in $\mathbb{Z}[x]$.
My approach, which hasn't been successful: I used that if $f(x)$ has a zero,$m$, in $\mathbb{Z}[x]$ then $m$ must divide $a_o$ = 1.
Two possibilities, $m=1$ & $m=-1$ $\to$ $f(x)= g(x)*(x-1)$ or $f(x) = g(x)*(x+1)$.
Then I tried dividing $f(x)$ by $(x-1)$ and $(x+1)$ both times I got a part left $\frac{p}{x\pm 1}$. I'm not sure if my approach was any good to begin with, and I'm also not sure if it was, how I'm supposed to proceed.
Any help would be tremendously appreciated.
 A: If $f$ has a rational root, then it is an integer factor of $p$, so is
$\pm1$ or $\pm p$. Each of these can be swiftly disposed of.
There is also the possibility of $f$ factoring into two quadratic polynomials. This factorisation will have the form $(x^2+ax+1)(x^2-ax+p)$
or $(x^2+ax-1)(x^2-ax-p)$. I'll leave it to you to figure out these cases.
A: A linear factor $x-a$ would require $a\mid p$, i.e., $a\in\{-p,-1,1,p\}$, and none of these is a root: $f(-p)=p^4\ne 0$, $f(-1)=p\ne 0$, $f(1)=p+2\ne 0$, $f(p)=p^4+2p\ne 0$.
Remains the case of two quadratic factors
$$f(x)=(x^2+ax+b)(x^2+cx+d).$$
Then $bd=p$ (constant term), $a+c=0$ (cubic term), $ad+bc=1$ (linear term),  and $0=b+ac+d$ (quadratic term).
Then $a(d-b)=ad+cb=1$, so $a=\pm1$, $ c=\mp1$.
This makes $b+d=1$ from the quadratic term, wheras $b+d=\pm(1+p)$ form the constant term and the possibly factorizations of $p$.
A: Split the problem into two cases:  $p = 2$ and $p \neq 2$.
If $p = 2$, consider reduction modulo $3$.  If $p \neq 2$, consider reduction modulo $2$.  
In both cases, it's easy to check that the polynomial has no roots over the finite field, but the possibility remains that the polynomial can factor into a product of two quadratics.  However, because there are a very limited number of irreducible quadratic polynomials over $\mathbb{Z}_2$ and $\mathbb{Z}_3$, it is relatively easy to determine that the polynomial is ultimately irreducible over the given finite field.  
We can then conclude that the polynomial is irreducible over $\mathbb{Q}$ (see Test 3 here).
