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Why is $x^2 +1$ is invertible in $Q[x]/(x^4 −3x^2 +6x)$?

Any ideas on how to prove this?

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By the Chinese remainder theorem one has that $\frac{\mathbb{Q}[x]}{(x^4 - 3x^2 +6x)} \cong \frac{\mathbb{Q}[x]}{(x)} \times \frac{\mathbb{Q}[x]}{(x^3 - 3x + 6)}$ is a product of fields, and the image of $x^2 + 1$ in each factor is nonzero.