prove that $f(x)=10x^4-18x^3+4x^2+7x+16 \in \mathbb{Z}[x]$ is irreducible in $\mathbb{Q}[x]$ I want to prove that $f(x)=10x^4-18x^3+4x^2+7x+16 \in \mathbb{Z}[x]$ is irreducible in $\mathbb{Q}[x]$.
If $f(x)$ can be analyzed in a product of two non-constant polynomials with coefficients $\in \mathbb{Q}$,then $f(x)$ can be analyzed in a product of two non-constant polynomials with coefficients $\in \mathbb{Z} $.
One case is: $f(x)=(ax^2+bx+c) \cdot (a'x^2+b'x+c')$.
So,to check if there are such polynomials,I have to solve the system:
$$\left\{\begin{matrix}
x^4: & 10=aa' \\ 
x^3: & -18=ab'+ba' \\ 
x^2: & 4=ac'+a'c+bb' \\ 
x: & 7=bc'+cb' \\ 
\text{ constant: } & 16=cc' 
\end{matrix}\right.$$
So,at the beggining,I suppose that $a=10,a'=1$ and then $a=5,a'=2$.
Then,do I also have to solve the system, with  $a'=10,a=1$ and $a'=5,a=2$ or is it the same??
 A: As a first step have fun checking out that the polynomial has no rational zeros. Leaving that to you [insert evil grin].
I am considering the possibility of $f(x)$ having two quadratic factors with integer coefficients. 
Let us project from the ring $\Bbb{Z}[x]$ to $\Bbb{Z}_2[x]$ by reducing all the coefficients modulo two. If $f(x)=g(x)h(x)$ for two quadratics $g(x),h(x)\in\Bbb{Z}[x]$, then modulo two we have
$$
\overline{g}(x)\overline{h}(x)=\overline{f}(x)=x.
$$
Because the ring $\Bbb{Z}_2[x]$ is a unique factorization domain, this means that one of $\overline{g}(x)$ and $\overline{h}(x)$ is $=x$ and the other $=1$.
In particular this implies that the leading coefficients of both $g(x)$ and $h(x)$ must be even for otherwise that factor would remain a quadratic after reduction modulo two as would then the product.
But this would imply that the leading coefficient of $f(x)=g(x)h(x)$ would be divisible by four, which is manifestly not the case. Thus such a factorization does not exist.
Q.E.D.
A: Here is another approach, that is nice, but is actually more tedious than the rest.
Suppose not. Let $ f(x) = g(x) h(x)$ where $g$ and $h$ are integer quadratic polynomials.
Observe that $ f(-15), f(-7), f(-5), f(-3), f(-1), f(1), f(3), f(7), f(25)$ are all prime.
WLOG, $g(x) = \{1, -1\}$ for 5 of those values. WLOG $g(x) = 1$ for 3 of those values. Hence, $g(x) = 1$ is a constant.
Note: This is actually more tedious, because you need to check for primality. I let the computer do that. It is likely easier to try and solve the Diophantine equation.
A: The polynomial $f(x)=10x^4-18x^3+4x^2+7x+16$ is irreducible iff $g(t)=f(t+1)$ is irreducible. We have
$$g(t)=10t^4+22t^3+10t^2+t+19.$$
There being no linear factors, the only factorization possible would be into
$$g(t)=(at^2+bt+c)(pt^2+qt+r).$$ Then $ap=10$ and without loss one may assume either $a=10,p=1$ or else $a=5,p=2.$ [possible negative choices could be removed since $-1$ could be factored out of all coefficients, and we may assume it is the first quadratic which begins with coefficient $10$ or $5$.
Now we also have that $cr=19,$ a prime. For each possible choice for the four coefficients $a,p,c,r$ we may use the $t^2$ coefficient of $10$, so that
$$ar+bq+cp=10,$$
and we may insert the values of $a,p,c,r$ and obtain the value of $bp$. In each case $bp$ is either a prime or a product of two primes, and one may check that the $t$ coefficient, which should be $1$, does not match. There are eight possible choices for the values of $(a,p,c,r)$ as follows, where the number after the four is the value of $bp.$
$(10,1,19,1;\ -19)$
$(10,1,-19,-1;\ 39)$
$(10,1,1,19;\ -181)$
$(10,1,-1,-19;\ 201)$
and four more starting out $(5,2,...)$.
In each case we can factor the value obtained for $bq$ and give assigned values to $b,q$ accordingly. By only checking the $t^1$ coefficient, which gives $br+cq=1$, it turns out that none of the possibilities gives this last equation. For example in the second case, where $bq=39=3\cdot 13$ we try $(b,q)=(39,1),(1,39),(13,3),(3,13)$ and none give the right value of $1$ to $br+cq.$ [In checking one may assume that say $b>0$ for setting up the cases, since a simultaneous sign change on $b,q$ only changes the sign of the quantity $br+cq$, and so we only need see if this ever comes out $\pm 1$.
This is admittedly a tedious computation, but at least it shows irreducibility.
