Irreducibility of Polynomials without Eisenstein In order to do a different task, I need to show that $f:=X^4-10X^2+1$ is irreducible over $\mathbb{Q}$ and that $g:=X^2Y^2-2Y^2-3X+6$ is irreducible over $\mathbb{Q}[X]$. Since I don't think that  Eisenstein criterion works here, I was wondering if there is another simple method to check that. Thanks in advance for any help!
 A: Hint:
For the first one, Suppose $f$ is reducible in $\mathbb{Q}[X]$. Clearly, there exists no rational root of the form $p/q$ (You could check this!!). So, it's possible that $f(x)$ is the product of two irreducible quadratics.
Suppose $f(X) = (X^2+aX+b)(X^2 + cX+d)$, where $a,b,c,d \in \mathbb{Q}$
Using this, you can arrive at a contradiction.
For the second one, as it has already been mentioned in the comments, $g$ is irreducible in $\mathbb{Q}[X,Y]$. Note that, you could use the fact that $\mathbb{Q}[X,Y]= (\mathbb{Q}[X])[Y]$. i.e Treat $g$ as a polynomial of $Y$ with coefficients in $\mathbb{Q}[X]$.
A: For the first one, there is a simplification due to https://en.wikipedia.org/wiki/Gauss%27s_lemma_(polynomial)
We see quickly that there are no rational roots. We are left with showing that the quartic is not the product of two quadratics with integer polynomials. The choices are
$$ (x^2 + Ax + 1)(x^2 + Bx + 1) $$ or
$$ (x^2 + Cx - 1)(x^2 + Dx - 1) .$$
From the coefficients of $x^3$ and $x$ being zero, we get $A+B = C+D = 0.$ Alright,
$$ (x^2 + Ax + 1)(x^2 -Ax + 1) $$ or
$$ (x^2 + Cx - 1)(x^2 -Cx - 1) .$$
All that is left is to get the coefficient $-10$ of $x^2,$ this may or may not be possible.
We reach $$   2 - A^2 = -10, $$
$$ -2 - C^2 = -10,  $$ for
$$ 12 = A^2, $$
$$ 8 = C^2 $$
Neither is possible with integers. You might try confirming that 
$$ (x^2 + x \sqrt {12} + 1)(x^2 -x \sqrt {12} + 1) = x^4 - 10 x^2 + 1.$$
$$ (x^2 + x \sqrt 8 - 1)(x^2 -x \sqrt 8 - 1) = x^4 - 10 x^2 + 1.$$
