Factoring $x^5+B x^4+C x^3+D x^2+E x+F=(x^2+a x+b)(x^3+p x+q)$ over $\mathbb{Q}$ For a quintic polynomial to be reducible to the following form over $\mathbb{Q}$:
$$x^5+B x^4+C x^3+D x^2+E x+F=(x^2+a x+b)(x^3+p x+q)$$
we need to match the coefficients ($a=B$ obviously, so we write the system of equations for the rest):
$$b+p=C \\ Bp+q=D \\ bp+Bq=E \\ bq=F$$
The system contains $4$ equations, so we get the condition on one of the coefficients of the original quintic. For convenience we choose $F$:

$$F=\frac{1}{2} \left(B^5-B^3C+3B^2D-2BE+CD \color{blue}{\pm} (B^3+D) \sqrt{(B^2-C)^2+4(BD-E)} \right)$$

$$b=\frac{1}{2} \left(B^2+C \color{blue}{\pm} \sqrt{(B^2-C)^2+4(BD-E)} \right)$$
$$p=\frac{1}{2} \left(C-B^2 \color{blue}{\mp} \sqrt{(B^2-C)^2+4(BD-E)} \right)$$
$$q=\frac{1}{2} \left(B^3-BC+2D \color{blue}{\pm} B \sqrt{(B^2-C)^2+4(BD-E)} \right)$$

For $b,p,q,F \in \mathbb{Q}$ we need:

$$E=BD+\frac{1}{4}(B^2-C)^2-n^2,~~~~~n \in \mathbb{Q}$$


My question is: Is the following statement correct?

A quintic polynomial in the form:
$$x^5+B x^4+C x^3+D x^2+E x+F,~~~~~B,C,D,E,F \in \mathbb{Q}$$
can be factored into $(x^2+a x+b)(x^3+p x+q)$ over $\mathbb{Q}$ iff:
$$E=BD+\frac{1}{4}(B^2-C)^2-n^2,~~~~~n \in \mathbb{Q}$$
$$F=\frac{1}{2} \left(B^5-B^3C+3B^2D-2BE+CD \color{blue}{\pm} 2(B^3+D)n \right)$$

So we get four free parameters $B,C,D,n$.
Or did I miss something?

Edit
If we allow $b,p,q$ to be irrational and/or complex, we can still keep $F$ rational by setting $D=-B^3$, which expands the set of quintics solvable this way.
 A: The statement is true.
You already proved the following : 

If a quintic polynomial in the form:
$$x^5+B x^4+C x^3+D x^2+E x+F,~~~~~B,C,D,E,F \in \mathbb{Q}$$
can be factored into $(x^2+a x+b)(x^3+p x+q)$ over $\mathbb{Q}$, then
$$E=BD+\frac{1}{4}(B^2-C)^2-n^2,~~~~~n \in \mathbb{Q}$$
$$F=\frac{1}{2} \left(B^5-B^3C+3B^2D-2BE+CD \pm 2(B^3+D)n \right)$$

In the following, let us prove the following : 

If $$E=BD+\frac{1}{4}(B^2-C)^2-n^2$$
$$F=\frac{1}{2} \left(B^5-B^3C+3B^2D-2BE+CD \pm 2(B^3+D)n \right)$$
  where $B,C,D,n\in\mathbb Q$, then a quintic polynomial in the form:
$$x^5+B x^4+C x^3+D x^2+E x+F$$
can be factored into $(x^2+a x+b)(x^3+p x+q)$ over $\mathbb{Q}$.

$$\begin{align}&x^5+Bx^4+Cx^3+Dx^2+Ex+F
\\&=x^5+Bx^4+Cx^3+Dx^2+\left(BD+\frac{1}{4}(B^2-C)^2-n^2\right)x
\\&\qquad +\frac{1}{2} \left(B^5-B^3C+3B^2D-2BE+CD\pm 2(B^3+D)n\right)
\\&=x^5+Bx^4+Cx^3+Dx^2+\left(BD+\frac{1}{4}(B^2-C)^2-n^2\right)x
\\&\qquad +\frac{1}{2} \left(B^5-B^3C+3B^2D-2B\left(BD+\frac{1}{4}(B^2-C)^2-n^2\right)+CD\pm 2(B^3+D)n\right)
\\&=x^5+Bx^4+Cx^3+Dx^2+\left(BD+\frac{1}{4}(B^2-C)^2-n^2\right)x
\\&\qquad +\frac{1}{4} \left(B^5+2B^2D-BC^2+4Bn^2+2CD\pm 4(B^3+D)n\right)
\\&=\left(x^2+Bx+\frac{B^2+C\pm 2n}{2}\right)\left(x^3+\frac{C-B^2\mp 2n}{2}x+\frac{B^3-BC+2D\pm 2nB}{2}\right)\end{align}$$
