Decomposing $x^4-5x^2+6$ over some fields My book asks me to decompose
$$x^4-5x^2+6$$
over:
$K = \mathbb{Q},\\ K = \mathbb{Q[\sqrt{2}]},\\ K = \mathbb{R}$
For $K = \mathbb{Q}$, I substituted $x² = a$ to get:
$$a²-5a+6 = (a-3)(a-2)$$
So Getting back to $a = x²$ we get:
$$x^4-5x^2+6 = (x²-3)(x²-2)$$
Also, for $\mathbb{R}$ we can simply factor $x²-3 = (x-\sqrt{3})(x+\sqrt{3})$ and $(x²-2) = (x-\sqrt{2})(x+\sqrt{2})$ so we have:
$$x^4-5x^2+6 = (x-\sqrt{3})(x+\sqrt{3})(x-\sqrt{2})(x+\sqrt{2})$$
but what about 
$$K = \mathbb{Q[\sqrt{2}]}\ ?$$

Your work is correct; for $\mathbb{Q}[\sqrt{2}]$, you surely have
$$
x^4-5x^2+6=(x-\sqrt{2})(x+\sqrt{2})(x^2-3)
$$
Is $x^2-3$ reducible over $\mathbb{Q}[\sqrt{2}]$? This would mean…
