<s>[INST] Strategy on factoring quartic polynomials: cross method. I found a method to factorize quartic polynomials which I don't understand how it works. 
It is presented like this:

Cross method
  This methodology lets to factorize ordered and completed 4th
  polynomials of the form:
$$F(x) = ax^4+bx^3+cx^2+dx+e$$
Rules:
  
  
*
  
*Factor the extreme terms with cross method to get a squared term (generally different from the squared term of the original
  polynomial). 
  
*Get $\Delta$ from the difference of the squared term of the polynomial and the term of the first step, and replace the result in
  the original polynomial. 
  
*Then, verify the binary combinations as double cross factoring.
  

I used for different exercises and it works, but I don't understand the basis for this method. 
Can anyone explain the reasons?
P.D.
Example
$$x^4+2x^3+3x^2+2x-3$$


*

*Factoring the extremes terms; $x^4$ and $-3$:


$$(x^2+3)(x^2-1)$$
The result of the cross method is $3x^2-x^2=2x^2$


*Calculating $\Delta$:


$$\Delta=\text{(original squared term)}-\text{(step one squared term)}= (3x^2)-(2x^2)=x^2$$
And replaced it in the original polynomial:
$$x^4+2x^3+x^2+2x-3$$


*Using the cross method for second and fourth term:


For second term: $(x^2+x)(x^2+x)$ is $2x^3$
For fourth term: $(x+3)(x-1)$ is $2x$
So, arranging the terms, the two factors are:
$$(x^2+x+3)(x^2+x-1)$$
 [/INST] I might be misunderstanding, but this just looks like concealed expanding $$(x^2+ax+b)(x^2+cx+d)\tag{1}$$ and comparing coefficients.


*

*Factoring extremes
Well, what we want is to choose $b$ and $d$ such that $(x^2+b)(x^2+d)$ matches leading and constant factor. What it amounts to is finding $b$ and $d$ such that $bd = -3$. Let's say we were lucky and chose the winning pair $b = -1,\ d=3.$


*Calculating $\Delta$
Why is this important? Well, $(x^2-1)(x^2+3) = x^4+2x^2-3$, so we are missing one $x^2$: $\Delta = 3x^2 - 2x^2$. 


*Covering for the missing $\Delta$
There is only one place where $\Delta$ can come from, that is from $a$ and $c$ - we must have $ac x^2 = \Delta$. Why? Well, if you expand $(1)$, you will find that $x^2$ term must be equal to $b + d + ac$. We found $b+d$ in step 1 and 2, so $acx^2$ is equal to $\Delta$. In this case $ac = 1$. There are two possibilities in $\Bbb Z$, either $a=c=1$, or $a=c=-1$. It is easy to decide which one, since it must fit the other terms as well, namely $x^3$ term is equal to $a + c$ and the $x$ term is equal to $ad + bc = 3a - c$. Thus, $a = c = 1$.


*Collecting terms
We simply write $(x^2+x-1)(x^2+x+3) = x^4+2x^3+3x^2+2x-3$ and feel the magic all around us.

Now, let us do the same thing, only transparently:
$$(x^2+ax+b)(x^2+cx+d) = x^4+(a+c)x^3+(b+d+ac)x^2+(ad+bc)x+bd$$
so we need to solve the following system in $\Bbb Z$:
\begin{array}{c r}\begin{align}
a+c&=2\\
b+d+ac&=3\\
ad+bc&=2\\
bd &= -3
\end{align} &\tag{2}\end{array}
There are two possibilities, either $b = 1$ and $d = -3$, or $b = -1$ and $d = 3$. We can see that former leads to no solution, so we will pursuit the latter, the system becomes:
\begin{align}
a+c&=2\\
ac&=1\\
3a-c&=2
\end{align}
The first and the third equation form linear system, which has unique solution $a = c = 1$, luckily, consistent with the second equation. Thus, we now know for certain that $$(x^2+x-1)(x^2+x+3) = x^4+2x^3+3x^2+2x-3.$$

Addendum:
There is a legitimate question why we would consider $(1)$ in the first case. Now, we are trying to factor a quartic over $\Bbb Z$. Let's say $p(x) = x^4 +2x^3+3x^2+2x-3$ and assume that $p = fg$. We have that $\deg p = \deg f + \deg g$, so there are two cases: $4=1+3$ or $4=2+2$. The first case would imply that $p$ has integer root and we can use rational root theorem  to rule that out. Thus, only case $4=2+2$ remains, which leads to considering $(1)$.
Finally, if the system  $(2)$ had no solutions in $\Bbb Z$, we would conclude that $p$ is irreducible over $\Bbb Z$.
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