# How can I show that $x^4+6$ is reducible over $\mathbb{R}$?

How can I show that the polynomial $x^4 + 6$ is reducible over $\mathbb{R}$ without explicitly finding factors?

I was trying to find a non-prime ideal that would generate it but I'm kind of lost as to how to proceed. Is there some sort of criterion that will allow me to show that it's reducible in $\mathbb{R}$?

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Sorry about that, it's not homework I'm just studying for a test. –  Danny May 15 '13 at 23:08
Let $a$ be positive. Thn $x^4+a=(x^2-kx+\sqrt{a})(x^2+kx+\sqrt{a})$ where $k=\sqrt{2\sqrt{a}}$. –  André Nicolas May 15 '13 at 23:32

Then use the fact that if a complex number $z$ is a root of a real polynomial, then $\bar z$ is also a root.
It follows that in $\Bbb R$ only the polynomials of degree one, and the quadratic polynomials with negative discriminants are the irreducible.
Wouldn't the fundamental theorem only allow me to show that the polynomial is reducible over $\mathbb{C}$? –  Danny May 15 '13 at 23:07
If $x=a+bi$ then $(x-a)^2=-b^2$. –  Berci May 15 '13 at 23:13