Showing that $x^4 -2x^2 +8 x+1$ is irreducible over $\Bbb Q$ I want to show that the polynomial
$$f(x)= x^4 -2x^2 +8 x+1$$
is irreducible over $\Bbb Q$.
I've proved it by a long method, but I need an easy and short method.
I've try to put $x=t+1$, but this fails.
My proof I reduce $f$ $\bmod 3$. I proved that there is no linear root by showing that $f(0), f(1), f(2)$ nonzero.
I proved that there is no quadratic root by checking all possible products of quadratics $x^ 2 + ax +b $.
 A: I'm not sure how much faster this is than the given method (showing that $f$ is irreducible modulo $3$), but here's a method that (as desired) avoids multiplying two general quadratics and solving the quadratic system in the coefficients, and instead only requires evaluating a finite number of a polynomial expressions (and a single polynomial long division):
By the Rational Root Theorem, the only possible rational roots of the given polynomial $f$ are $\pm 1$, and substituting shows that neither of these are solutions, so the polynomial has no linear factors. Thus, if it is reducible, it is a product of two irreducible quadratics, and hence its factorization modulo any prime has a quadratic factor (not necessarily irreducible, of course).
If we reduce the polynomial modulo $5$, substituting shows that $x = 2$ is a root, and then polynomial long division gives that
$$f(x) \equiv (x + 3)(x^3 + 2 x^2 + 2 x + 2) \bmod 5.$$
Now, if the cubic factor is reducible modulo $5$, it has a linear factor modulo $5$, but computing $f(0), \ldots, f(4)$ (and observing that each is nonzero) shows that it does not. Therefore, the cubic is irreducible, and in particular $f$ does not have a quadratic factor modulo $5$; thus, by the previous paragraph, it is irreducible.
A: We will use a theorem that states that if $f$ is a polynomial:$f(x)\inℤ[x]$, then $f(x)$ can be analysed as the product of two polynomials with lesser degrees, $r$ and $t$ in $ℚ[x]$ iff such an analysis-with polynomials of the same degrees, $r$ and $t$, exists in $ℤ[x]$.
From this, it follows that if $f(x)=x^n+a_{n-1}x^{n-1}+...+a_0 \inℤ[x]$ and $a_0≠0$, and if $f(x)$ has a root in $ℚ$, then it also has a root $m$ in $ℤ$, and $m$ must divide $a_0$.
So for $f(x)=x^4-2x^2+8x+1$, if $f(x)$ has a linear factor in $ℚ[x]$, then it has a root in $ℤ$ and this root must be a divisor of $1$ (your $a_0$). So it must be either $1$ or $-1$. But $f(1)=8$ and $f(-1)=-8$, so such an analysis does not exist.
If $f(x)$ has two quadratic factors in $ℚ[x]$, then it can be factored also in $ℤ[x]$ as follows: $(x^2+ax+b)(x^2+cx+d)$
Now you can easily compare the terms of $f$ with the given expression(solve the system for ($a,b,c,d$) and see that such an expression does not exist.
A: $f(x)=x^4-2x^2+8x+1=(x^2+ax+b)(x^2+cx+d$) implies inmediately $a+c=0; d=1/b$ $\implies\frac{1}{b}+b=-2$ $\iff(b+1)^2=0$. Also $a(\frac{1}{b}-b)=8$ which  implies
0 =8, absurde. Thus there are not quadratic factors.On the other hand a linear factor must be monic and does not be other than (x$\pm 1$) which is easily seen to be impossible (hence not a cubic factor).
