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.
