Show that $x^4-10x^2+1$ is irreducible over $\mathbb{Q}$ How do I show that $x^4-10x^2+1$ is irreducible over $\mathbb{Q}$? Someone says I should use the rational root test, but I don't exactly know how that applies. Thanks for any input. 
 A: Hint: If $p(x)$ is a factor, $p(-x)$ is a factor. Show that there can be no linear factor, and no quadratic factor of the form $x^2-a$. Then try to solve $x^4-10x^2+1 = (x^2+ax+b)(x^2-ax+b)$.
A: Perhaps it is worth noting that $x^{2n} -10 x^n + 1$ is irreducible over $\mathbb{Q}$ for every natural $n\ge 1$. Indeed, for $n=1$ the polynomial 
$$x^2 - 10 x + 1 = (x-(5 + 2\sqrt{6})\,)(x-(5-2\sqrt{6})\,)$$
is irreducible over $\mathbb{Q}$.  Now, it is enough to check that in the decomposition
$$x^{2n}-10 x^n + 1 =(x^n-(5 + 2\sqrt{6})\,)(x^n-(5-2\sqrt{6})\,)$$
both of the factors are irreducible in $\mathbb{Q}(\sqrt{6})[x]$.  
Let's consider the general case of a binomial $x^n - \alpha$ over some field $F$. Decompose into linear factors over some extension 
$x^n - \alpha = (x-\gamma_1)\cdot \ldots \cdot (x-\gamma_n)$. Assume that $x^n -\alpha$ is reducible over $F$. Then we can find $k$ factors ($1\le k \le n-1$) whose product is in $F[x]$. Assume wlog that $$(x-\gamma_1)\cdot \ldots \cdot (x-\gamma_k)\in F[x]$$   We conclude 
$\gamma_1 \cdot \ldots \cdot \gamma_k = \beta \in F$.  Now raise to the power $n$ and get $\beta^n = \gamma_1^n\cdot \ldots \cdot \gamma_k^n  = \alpha^k$.  Conclusion:
If $x^n - \alpha$ is reducible in $F[x]$ then there exists $1\le k \le n-1$ and $\beta \in F$ so that 
$$\alpha^k = \beta^n$$
Back to our problem. It is not hard to check that the group of units of the ring of integers of the field $\mathbb{Q}(\sqrt{6})$ is generated by $5 + 2\sqrt{6}$ and $-1$. Therefore,  for $\alpha = 5\pm 2 \sqrt{6}$ we cannot have an equality as above. 
A: We have that $p(x)=x^4-10x^2+1$ splits as $(x^2-2)(x^2+2)$ over $\mathbb{F}_5$, hence $p(x)$ has no linear factors over $\mathbb{Q}$ and by assuming it splits over $\mathbb{Q}$, since $p(x)=p(-x)$, it must satisfy:
$$ x^4-10x^2+1 = (x^2+ax+b)(x^2-ax+b) $$ 
with $b^2=1$ and $a^2-2b=10$, contradiction.

With a little reverse-engineering, I guess you just want to show that your polynomial is the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb{Q}$. We have that $\mathbb{Q}[x,y]/(x^2-2,y^2-3)$ is a vector space over $\mathbb{Q}$ with dimension $4$ and a base given by $1,x,y,xy$. With respect to such a base:
$$\begin{pmatrix}1 \\ (x+y) \\ (x+y)^2 \\ (x+y)^3 \\ (x+y)^4\end{pmatrix}= \begin{pmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 5 & 0 & 0 & 2 \\ 0 & 9 & 11 & 0\\49& 0 & 0 & 20 \end{pmatrix}\begin{pmatrix}1 \\ x \\ y \\ xy \end{pmatrix}$$
and by applying Gaussian elimination we get that the fifth row of the last matrix can be expressed as a linear combination of the previous four rows, so that $z^4-10z^2+1$ is a polynomial in $\mathbb{Q}[z]$ that vanishes at $z=\sqrt{2}+\sqrt{3}$. Since the rank of such a matrix is four (easy to check), the last polynomial is also the minimal polynomial of $\sqrt{2}+\sqrt{3}$, so it is irreducible.
A: There are so many ways to see the irreducibility of the polynomial $p(x)=x^4-10x^2+1$. Adding yet another way, using Eisenstein and trickery.
We see that 
$$
r(x):=p(x+1)=x^4+4x^3-4x^2-16x-8.
$$
At this point users familiar with $2$-adics and Newton's polygon would already see a way. A more elementary route is to observe that $r(x)$ is irreducible over the rationals if and only if $r(2x)$ is. But
$$
r(2x)=16x^4+32x^3-16x^2+32x-8=8 q(x)
$$
with
$$
q(x)=2x^4+4x^3-2x^2+4x-1.
$$
The other trick we need is to use the fact that a polynomial (with a non-zero constant term) is irreducible iff its reciprocal polynomial
$$
\tilde{q}(x)=x^4q(\frac1x)=-x^4+4x^3-2x^2+4x+2
$$
is irreducible. Here Eisenstein with $p=2$ kicks in.
A: The rational root theorem gives you a list of possible rational roots of your polynomial.  Can you make that list for your polynomial?  Once you try them all, you know that there is no factorization $x^4-10x^2+1=(x-r)(q(x))$ for some $r \in \Bbb Q$  and polynomial $q(x)$.  You still need an argument that there is not factorization into two quadratics with coefficients in $\Bbb Q$
A: You can use the Generalized Arthur Cohn's irreducibility criterion (no other answer has mentioned it yet). But the solution is ugly -- you have to prove the primality of an integer larger than $10^9$.
In this case, $x^4-10x^2+1$ has a term with a negative coefficient, so the criterion couldn't be directly applied.
If a term with a negative coefficient had an odd exponent, then the substitution $x=-y$ might help. In this case, the exponent is even.
Trying substitutions $x=\pm y\pm 1$, $x=\pm y\pm 2$, $x=\pm y\pm 3$ doesn't work ($12$ tries overall, or $6$, because $|y+1|=|-y-1|$, $|y-1|=|-y+1|$, etc., and the exponents are all even) for giving a polynomial with only positive coefficients. Finally, $x=y+4$ does work, giving us the polynomial $y^4+16 y^3+86 y^2+176 y+97$.
If $y>176$ is an odd integer, then the polynomial is even and larger than $2$, so not prime. So $y$ must be even if we want the polynomial to be prime.
Trying $y=178$ doesn't give a prime number (you could prove it's divisible by $23$), but $y=180$ does, so we're done. Clearly proving the primality by hand would be a problem, so it's an ugly solution.
By a Gauss's Lemma for polynomials irreducibility over $\mathbb Z[x]$ for a polynomial with integer coefficients implies irreducibility over $\mathbb Q[x]$ (the Wikipedia page for Cohn's criterion currently uses $\mathbb Z[x]$).
A: $$f(x)=x^4-10x^2+1=(x^2+2\sqrt2x-1)(x^2-2\sqrt2x-1)$$ Hence $f(x)$ has $4$ irrational roots. It is enough to know $f(x)$ is irreducible over $\mathbb Q$.
A: This can be seen from the following fact:

If $P\in{\bf Z}[X]$ is monic with $d=\deg(P)\geqslant1$ and takes values in the set $\{\pm 1\}\cup\{\pm p\mid p\text{ prime}\}$ in at least $2d+1$ distinct integers $a_1,\dots,a_{2d+1}\in{\bf Z}$, then it is irreducible over ${\bf Q}$.

For the proof, see Theorem 5.1. The example of this polynomial is also given, where the integers $a_1,\dots,a_9$ can be taken to be $0,\pm2,\pm4,\pm6,\pm8$.
As a side note, this polynomial is reducible mod every prime (adding it because the question comment's link is dead).
