Does there exist a factorization consisting of polynomials of $x^4+3x^2+6$

Question

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?

Answer 1

$$x^4+3x^2+6= x^4+2\sqrt6x^2+6-(2\sqrt6-3)x^2=$$ $$=\left(x^2-\sqrt{2\sqrt6-3}x+\sqrt6\right)\left(x^2+\sqrt{2\sqrt6-3}x+\sqrt6\right)$$

Answer 2

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.