Suppose I have a polynomial $A$, which I factorize as $A=BC$ (where $B$ and $C$ are polynomials with integer, rational or real coefficients). When factoring $B$ as $B=B'$ and $C$ as $C=C'$ (to irreducibly polynomials over the respective field) does $A=B'C'$ gives the full factorization.
In other words, does factorizing the parts of a partially factorizing polynomials gives the full factorization of the original polynomial (over all of integers, rationals or reals)?
Regards and thanks.
On
First off, note that polynomials over a field form a unique factorization domain, and polynomials over a unique factorization (such as $\mathbb{Z}$) also form a unique factorization domain. In a unique factorization domain the irreducible factors are the same as prime factors, and the prime factors are unique (up to their order and multiplication by units). (This would be true even if we were dealing with polynomials in more than one indeterminate.)
Suppose $A=BC$. Any irreducible factor of $B$ is an irreducible factor of $A=BC$. (Verify that a factor of $B$ is a factor of $BC$, and irreducibility of a polynomial is intrinsic with respect to the given ring of polynomials.)
Given the uniqueness of prime factors (aka irreducible factors in this context), it is not possible to "get stuck" or make a wrong choice of top-level factors in searching for the irreducible ones.
As long as you stick to one of either integers, rationals, or reals, you can't go astray by choosing a "wrong" factorization. The rings $\mathbb Z[X]$, $\mathbb Q[X]$ and $\mathbb R[X]$ are all _unique factorization domains, so if you have $A=BC$ in one of these rings, then you know that the irreducible factors if $A$ are the irreducible factors of $B$ together with the irreducible factors of $C$.
On the other hand, if you mix and match base rings -- say, is $A$ is an integer polynomial and $B$ and $C$ are real polynomials, then there may be irreducible factors of $A$ in $\mathbb Z[X]$ that are not factors of $B$ and $C$. For example, this would be the case if $A$ is irreducible over the integers, and $B$ and $C$ are factors of lower (but nonzero) degree.
However, since $\mathbb Q$ is the fraction field of $\mathbb Z$, you do get to mix and match between those. The irreducible polynomials over $\mathbb Q$ are exactly the irreducible nonconstant polynomials over $\mathbb Z$ (times rational constants). So it is possible to factor integer polynomials by considering factors with rational coefficients, if you take care about handling degree-0 factors separately.