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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?

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    yes it does, compute the complex zeros – 2017-02-08
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    $$ \left( x^2 + \frac{3}{2} \right)^2 + \frac{15}{4} $$ – 2017-02-08
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    An algorithmic approach: Suppose $x^4+3x^2+6 = (x^2+ax+b)(x^2+cx+d)$. Expand the right-hand side and show that this forces $a=-c$ and $b=d$. You're left with a system of two equations for the two remaining unknowns. – 2017-02-08
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    @AshwinIyengar that's true...am thinking about it... but how to go on then? I mean, you'll definitely not be able to factor it in the way $(x-x_i)\ldots$ – 2017-02-08
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    gp and *Mathematica* probably think you're looking for a factorization with *integer* coefficients. The theorem only guarantees factors with *real* coefficients. – 2017-02-08

2 Answers 2

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

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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.