Factorization $x^4+px^3+qx^2+r x +s=(x^2+a x +b)(x^2+\bar a x +\bar b)$ Question:
Under what condition, does the quartic polynomial with rational coefficients
$p$, $q$, $r$ and $s$ factorizes as
$$x^4+px^3+qx^2+r x +s= (x^2+a x +b)(x^2+\bar a x +\bar b) $$
with $a$, $b$ complex numbers, along with their conjugates $\bar a $, $\bar b$.
Examples:
$$x^4+2x^3+6x^2+2x+1=( x^2 +(1-i \sqrt3)x +1) (x^2 +(1+i \sqrt3)x +1) $$
$$x^4+2x^3+4x^2+2=( x^2 +(1+i)x +(1-i)) (x^2 +(1-i)x +(1+i))  $$
Note that the symmetry of coefficients leads to such factorization, as seen in the first example; but not exclusively so, as shown by the second example.
Is there any test on the coefficients $p$, $q$, $r$ and $s$ that can be carried out to determine the possibility of such factorization? I reviewed here the discriminate tests on the nature of roots for quartic equations and did not find anything applicable.

Short answer:
Actually, the things I do, do not involve any special manipulation. It just contains simple algebra.
Let $a,b,c,d \in\mathbb R$  then we have,
$$\left(x^2+(a+bi)x+(c+di)\right)\left(x^2+(a-bi)x+(c-di)\right)=x^4+2ax^3+(a^2+b^2+2c)x^2+(2ac+2bd)x+(c^2+d^2).$$
$$x^4+px^3+qx^2+rx+s=x^4+2ax^3+(a^2+b^2+2c)x^2+(2ac+2bd)x+(c^2+d^2)$$
which follows
$$\begin{cases} p=2a \\ q=a^2+b^2+2c \\ r=2ac+2bd \\s=c^2+d^2\end{cases} $$

*

*If $d=0$, then the system of equation becomes extremely simple. I'll leave this case to you.  Here we will work with $d≠0.$


*If $d≠0$, then $s-c^2≠0$. We have,
$$\begin{cases} a=\frac p2 \\b^2+2c=q-\frac{p^2}{4} \\pc+2bd=r \\c^2+d^2=s \end{cases}$$
$$\implies \begin{cases}\left( \frac{r-pc}{2d}\right)^2+2c-q+\frac{p^2}{4} =0 \\ c^2+d^2=s \end{cases}$$
Finally we get,
$$\frac{(r-pc)^2}{4(s-c^2)}+2c-q+\frac{p^2}{4} =0$$
$$\color {gold}{\boxed {\color{black}{8 c^3-4qc^2+(2pr-8s)c+(4qs-sp^2-r^2)=0.}}}$$
As it seems, we obtain the cubic equation with respect to $c$.
If all of the coefficients of the cubic equation are real numbers, then it has at least one real root. Therefore, we can always choose $c$ to be a real number.
If you accept $a,b,c,d$ as rational, then we can immediately use the Rational Root theorem.
For this, all we need is to find the factors of this expression $\color{red}{\dfrac{4qs-sp^2-r^2}{8}.}$
We're almost done.  We have the last two restrictions.
After finding real $c$, then we must check the following two cases:

*

*$s-c^2≥0$


*$p^2-4q+8c≤0.$
If these conditions hold, then you can easily find the $a,b,d$ from the system of equations.
End of the answer.
