I would like to see if I'm right about these polynomials I tried to prove are irreducible:
1)
For the first polynomial I used that if a polynomial is irrational over $\mathbb{Z}_p$ for $p$ prime, and $p$ does not divide the $x^n$ coefficient of this polynomial (suppose degree $n$), then it is irreducible over $\mathbb{Q}$.
$$p(x) = x⁴+2x³+2x²+2x+2$$
I choose $\mathbb{Z}_2$. Well, over $\mathbb{Z}_2$ we get:
$$p(x) = x⁴$$
the only possible root for this polynomial is $x=0$, because $1⁴=1$. Can I say this polynomial is irreducible because its only root is $0$ and I would break it into $(x-0)(x-0)(x-0)(x-0)$ which is just $x⁴$?
2) For the second polynomial I used the Eisenstein's criteria: if $p$ is a prime such that $p²$ does not divide $a_0$, but not $a_n$ and $p$ divides the other coefficients, then $p(x)$ is irreducible over $\mathbb{Q}$:
$$p(x) = x^7-31$$
it's clear that $p=31$ will divide $a_0$ but $p²$ will not, and also $p$ does not divide $a_n$. In fact, every polynomial in the form $x^n-p=0$ will be irreducible over $\mathbb{Q}$ by this criterion
3)
$$p(x) = x^6+15$$
We have that $p=3$ will divide $a_0=15$ but $p^2=9$ will not. Also, $p=3$ will not didivde $a_n=1$, so $p$ is irreducible over $\mathbb{Q}$.
4)
$$p(x) = x^3+6x^2+5x+25$$
I cannot use the criterion here because $p$ must divide $25$ so $p=5$ but $p$ will not divide $a_2 = 6$. So I'll try to reduce it $mod p$. Taking it $mod 5$ I get:
$$p(x) = x^3+x^2$$
well, it didn't work. So I'll try $mod 3$, we get:
$$p(x) = x^3+2x+1$$
which is reducible…
So let's try $mod 2$:
$$p(x) = x^3+x+1$$
which is still reducible :c
Any ideas on this one?
5) $$p(x) = x^4+8x^3+x^2+2x+5$$
if we reduce mod $2$ we get:
$$x^4+x^2+5$$
which I suspect is irreducible. It's always positive, so it has no roots, so it can't be factored into at least one $1$ degree factor. The only possibility would be to factor it in two irreducible $2$ degree polynomials, which I'll try later to prove. Wolfram Alpha didn't give its factors so I think it's a signal.
6)
$$p(x) = x^4+10x^3+20x^2+30x+22$$
reduction mod $2$ would be a good idea, but we have that $2$ divides $a_0$ but $2^2$ will not divide it. $2$ divides every other coefficients except $a_n=1$ so this polynomial is irreducible
Best Answer
(1) is screaming for you to use eisenstein's criterion since everything has a factor of $2$ except the first term.
(4) is irreducible mod $2$, if you plug in either $0$ or $1$ we get a value not congruent to $0$ mod $2$.
For (5) yes you should check to see whether it factors as 2 quadratic polynomials, and whether it has a rational root. Or just find a $p$ such that it is irreducible in $\mathbb{Z}/(p)$ and check that it does not factor as the product of two quadratic polynomials.