Show that $$x^4+n(2-n)x^2+n^2$$ reducible over Q for any natural number n.
Here is what I did $$x^4+n(2-n)x^2+n^2=x^4+2nx^2-n^2x^2+n^2=x^4+2nx^2+n^2-n^2x^2$$ $$=(x^2+n)^2-(nx)^2$$ $$=(x^2+n-nx)(x^2+n+nx)$$
Show that $$x^4+n(2-n)x^2+n^2$$ reducible over Q for any natural number n.
Here is what I did $$x^4+n(2-n)x^2+n^2=x^4+2nx^2-n^2x^2+n^2=x^4+2nx^2+n^2-n^2x^2$$ $$=(x^2+n)^2-(nx)^2$$ $$=(x^2+n-nx)(x^2+n+nx)$$
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