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.

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.

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.

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 $x^4-10x^2+1 = (x^2+ax+b)(x^2-ax+b)$.

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 $\Bbb Q$

$$f(x)=x^4-10x^2+1=(x^2+2\sqrt2x-1)(x^2-2\sqrt2x-1)$$ Hence $f(x)$ has $4$ irrational roots. It is enough to know $f(x)$ is irreducible over $\mathbb Q$.