Simple Generalized Quadratic Factorization

Important in my research is a simple generalization of the factorization of a quadratic. That quadratic is easily considered in the complex plane.

In the complex plane:

P(x) = (g₁(x) + 1)(g₂(x) + 2)

where P(x) is a primitive quadratic with integer coefficients, g₁(0) = g₂(0) = 0, but g₁(x) does not equal 0 for all x.

The simplest example is: P(x) = x² + 3x + 2

Solving for g's in general is easily done with some substitutions, where one will seem superfluous, but is important. To solve with my approach we will need a new variable k, and two new functions.

Introduce k, where k is a nonzero, and new functions f₁(x), and f₂(x), where:

g₁(x) = f₁(x)/k and g₂(x) = f₂(x) + k-2

Multiply both sides by k, and substitute for the g's, which gives me the now symmetrical form:

k*P(x) =  (f₁(x) + k)(f₂(x) + k)

The purpose of forcing symmetry is to allow for solutions where also need one more function, then can solve for the f's using the quadratic formula.

And introduce H(x), where: f₁(x) + f₂(x) = H(x)

Then just solve for one of the f's, and substitute so I can find:

f₁²(x) - H(x)f₁(x) - kH(x) - k² + k*P(x) = 0

And then you can solve for f₁(x) using the quadratic formula:

f₁(x) = (H(x) +/- sqrt[(H(x) + 2k)² - 4k*P(x)])/2

And H(x) is a handle for every possible factorization with the g's, and has a key constraint: H(0) = -k + 2.

If wished, as an exercise reader can confirm our simple factorization solution with:

k = 1, H(x) = 2x+1, and P(x) = x² + 3x + 2

Which will give x and x+1 as solutions for the f's, and x as a solution for the g's. And, a curious reader might next consider what happens if you force a non-polynomial factorization, for instance with H(x) = 3x + 1. Notice you only need make sure H(0) = 1.

Have been considering in the complex plane, but notice can easily move to rings by how control variables. For example if k is an integer, and is unit, which means 1 or -1, and H(x) is an algebraic integer function, then the f's are forced to be as well. And also the g's are so forced.

So we know that we can get every possible algebraic integer factorization using our generalized approach, which is important.

However, we can also choose to pick some other k, like k = 2, or k = 3, Which is of mathematical interest beyond the scope of this instructional post.

Have introduced a generalized quadratic factorization with a simple quadratic, and stepped through a basic approach to finding solutions for that factorization in the complex plane.


James Harris