<s>[INST] Does there exist a factorization consisting of polynomials of $x^4+3x^2+6$ So as I understand it every polynomial with real coefficients should have a factorization consisting of polynomials of degree one(In case of real roots) and degree two(complex roots).
But I have been unable to find such a factorization of the polynomial $x^4+3x^2+6$. Even using gp and Mathematicas Factor function I only get the original polynomial as an answer. 
So my question is. Does there exist a factorization?
 [/INST] Let $y=x^2$ then solving the quadratic $y^2+3y+6=0$ gives $y = \frac{1}{2}(-3 \pm i \sqrt{15})$.
Then solving:
$$
x^2 = \frac{1}{2}(-3 + i \sqrt{15}) \implies
x_{1,2} = \pm \frac{1}{2}\left(\sqrt{2\sqrt{6}-3}+i \sqrt{2\sqrt{6}+3}\right) \\
x^2 = \frac{1}{2}(-3 - i \sqrt{15}) \implies
x_{3,4} = \pm \frac{1}{2}\left(\sqrt{2\sqrt{6}-3}-i \sqrt{2\sqrt{6}+3}\right)
$$
Note that $x_3=\overline{x_1}$ and $x_4=\overline{x_2}$ so both polynomials $(x-x_1)(x-x_3)$ and $(x-x_2)(x-x_4)$ have real coefficients, which can be easily calculated once you know the roots. Those two polynomials are the quadratic factors of the original polynomial.
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