Factoring of polynomial over the rationals (quick question)

Question

Okay this might be a stupid question, but couldn't find any clear explanation on the internet, so here goes.

If I factor a polynomial into products of irreducible polynomials, are the products of these irreducible polynomials then again an irreducible polynomial?

For clarification, here is the problem im working on:

Let $J$ be the ideal of $\mathbb{Q}[x]$ generated by two polynomials: $f(x)=(x+1)(x+2)(x^2 +1)$ $\textit{and}$ $g(x)=(x+1)^2 (x+2)^2.$ Is the ideal $J$ prime? Is it maximal? Is it principal?

Clearly the common factors of $f$ and $g$ are $(x+1)(x+2)$, but would it be right to say that $p(x)=(x+1)(x+2)$ is an irreducible polynomial over $\mathbb{Q}$?

If this is the right approach, I think I can handle the rest of the problem myself.

Thanks in advance :)

Answer 1

If I factor a polynomial into products of irreducible polynomials, are the products of these irreducible polynomials then again an irreducible polynomial?

No, quite the opposite in fact. In fact, that's basically what irreducible means: $f$ is irreducible if $f=gh$ implies one of $g,h$ are constant, i.e. there's no nontrivial factorization. So if you take irreducible $g,h$ (which requires them to be have degree at least one) and multiply them together, you get a polynomial $f$ whose factorization is $gh$ .

Think of it like prime numbers - irreducible polynomials are like primes. Multiply two primes, you get a composite number, never, ever a prime number.

Clearly the common factors of $f$ and $g$ are $(x+1)(x+2)$, but would it be right to say that $p(x)=(x+1)(x+2)$ is an irreducible polynomial over $\mathbb{Q}$?

It would not be irreducible by the previous discussion as a result.

Answer 2

Simply $p(x)$ is not irreducible, because you have exhibited a factorisation. The factorisation you have given is a factorisation of $p(x)$ into irreducible factors.