I need to find a factorization of both $f_{1}=2x^{2}+4x+6$ and $f_{2}=2x^{2}+4x-6$ into a product of irreducible elements of $\mathbb{Q}[x]$.
I already was able to do so in the case of $\mathbb{Z}[x]$:
- $f_{1}=2(x^{2}+2x+3)$
- $f_{2}=2(x-(-3))(x-1)$
Note that for $f_{1}$, $x^{2}+2x+3$ has no rational roots (I tested each of the possible combinations of $\frac{p}{q}$ where the possibilities for $p$ are $\pm 1$, $\pm 3$ and the possibilities for $q$ are $\pm 1$.
I was wondering, therefore, how factoring these polynomials over $\mathbb{Q}$ would be different (and I guess, essentially, how it would proceed)? Especially in light of the fact that all integers are rational, no nonintegral rationals came up as I was factoring over the integers, and the result that states that a polynomial is irreducible over $\mathbb{Z}$ iff it is irreducible over $\mathbb{Q}$.
Thanks in advance for your time and patience.