I want to show that x^{4}-22x^{2}+1 is irreducible over Q. I believe I need to use the Eisenstein criterion, but Im not really sure how.
Or you could just do this: Let $\displaystyle \displaystyle \begin{align*} X = x^2 \end{align*}$ which makes your equation
$\displaystyle \displaystyle \begin{align*} x^4 - 22x^2 + 1 &= X^2 - 22X + 1 \\ &= X^2 - 22X + (-11)^2 - (-11)^2 + 1 \\ &= (X - 11)^2 - 120 \\ &= \left( X - 11 - 2\,\sqrt{30} \right) \left( X - 11 + 2\,\sqrt{30} \right) \\ &= \left( x^2 - 11 - 2\,\sqrt{30} \right) \left( x^2 - 11 + 2\,\sqrt{30} \right) \\ &= \left[ x^2 - \left( \sqrt{ 11 + 2\,\sqrt{30} } \right) ^2 \right] \left[ x^2 - \left( \sqrt{ 11 - 2\,\sqrt{30} } \right) ^2 \right] \\ &= \left( x - \sqrt{ 11 + 2\,\sqrt{30} } \right) \left( x + \sqrt{ 11 + 2\,\sqrt{30} } \right) \left( x - \sqrt{ 11 - 2\,\sqrt{30} } \right) \left( x + \sqrt{ 11 - 2\,\sqrt{30} } \right) \end{align*}$
Clearly none of these linear factors is rational, so your quartic can not be reduced over the rationals.
Hi Prove It,
I don't quite follow your argument (at least the last few lines).
$\displaystyle x^4-5x^2+6=(x-\sqrt2)(x+\sqrt2)(x-\sqrt3)(x+\sqrt3)$ with none of the linear factors in Q[x]. So the
polynomial is irreducible over the rationals? But
$\displaystyle x^4-5x^2+6=(x^2-2)(x^2-3)$ and so the polynomial is reducible in Q[x].