$x^4+x^3+x^2+x+1$ irreducible over $\mathbb F_7$ This question came from the answer here.
The answer claims that $x^4+x^3+x^2+x+1$ is irreducible over $\mathbb F_7$. I can check that it has no roots in the field, but why can't it be written as a product of two quadratics? I tried the method of undetermined coefficients: $$x^4+x^3+x^2+x+1=(x^2+ax+b)(x^2+cx+d)=\\ x^4+(a+c)x^3+(ac+b+d)x^2+(bc+ad)x+bd$$ so $$a+c=1\\ac+b+d=1\\bc+ad=1\\bd=1$$ but I wasn't able to solve this. Maybe there is a way to see irreducibility by only using general finite fields theory?

Yet an other solution. It is based on the fact that 
$$ f = x^4+x^3+x^2+x+1\in\Bbb F_7[x] $$
is reciprocal, and tries to use the idea in the OP. (There should be no splitting in two polynomials of degree two.)
Assume there is a factorization
$$
f=(x^2+ax+b)(x^2+cx+d)\ ,\qquad a,b,c,d\in\Bbb F_7\ .$$
Let $u\ne 0$ be a root of the first factor in some extension (of degree two). Then $1/u$ is also a root. We distinguish two cases. 
First case: $1/u$ is also a root of the first factor. Then $b=1$, then $d=1$, $c+a=1$, the product 
$$(x^2+ax+1)(x^2+(1-a)x+1)$$
is reciprocal, 
and we have to look for a match of the coefficient in $x^2$, for a few values of $a$ (modulo the action $a\leftrightarrow (1-a)$). There is no such $a$ with $1+a(1-a)+1=1$.
Second case: $1/u$ is not a root of the first factor. Let $u$, $v$ be the two roots of $x^2+ax+b$. Then $1/u$, $1/v$ are the roots of the second factor, which is the reciprocal polynomial, made monic, so we expect an equality of reciprocal polynomials
$$
(x^2+ax+b)(1+ax+bx^2) = b(x^4+x^3+x^2+x+1)\ .
$$
The two equations obtained, $ab+a=b$ and $1+a^2+b^2=b$, have no solution in $\Bbb F_7$. (The substitution of $a=b/(b+1)$ in the second equation leads to... $(b^4+b^3+b^2+b+1)/(b+1)^2=0$.)

Note: This is arguably shorter then taking all possible polynomials $x^2+ax+b$, $b\ne 0$, associating $c=1-a$, $d=1/b$, and checking if the other conditions among $a,b,c,d$ are satisfied.
Note: Computer algebra programs show that we have an irreducible polynomial in our hands. For instance using sage:
sage: R.<x> = PolynomialRing(GF(7))
sage: f = x^4 + x^3 + x^2 + x + 1
sage: f.is_irreducible()
True

One can use then computer power also as follows to conclude. (With a slightly bigger effort than for the typing, one can also conclude humanly.) Assume $f$ splits in two (or more factors). One can check there is no root in $\Bbb F_7$. Then there is a root in $\Bbb F_{7^2}=\Bbb F_{49}$, so the polynomials $f$ and $x^{49}-x$ have a common divisor. The gcd is but... we type
sage: gcd( x^49 - x, f ) 
1
sage: gcd( x^(7^4) - x, f ) 
x^4 + x^3 + x^2 + x + 1

For this, one can also compute easily $(x^{49}-x,f)$ as a human by noting that $f$ divides $x^5-1$, so $(x^{49}-x,f)=(x^{49}-x^4+x^4-x,f)=(x^4-x,f)=(x^3-1,f)=(x^2+x+1,f)=(x^2+x+1,x+1)=(1,x+1)=1$. 
