Showing that a polynomial is irreducible

Question

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.

Answer 1

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 $p$.

Answer 2

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.

Answer 3

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).