I'm trying to understand how to check if polynomials are irreducible.
For example: Are these polynomials irreducible over $\mathbb{Q}$? $$f(x)=x^3+4x^2+3\\g(y)=12y^4+5y^3-10y^2+25y+10$$
A rational root $\frac{r}{s}$ for $f(x)=x^3+4x^2+3$ gives $$s|1 \Rightarrow s=\pm 1\\ r|3 \Rightarrow r=\pm 1, 3\\ Possible\ roots:\frac{3}{1},-\frac{3}{1},\frac{1}{1},-\frac{1}{1}$$
None of the possible roots gives $f(x)=0$. Therefor $f(x)$ is irreducible.
A rational root $\frac{r}{s}$ for $g(y)=12y^4+5y^3-10y^2+25y+10$ gives $$s|12\Rightarrow s=\pm 1,2,3,4,6,12\\ r|10 \Rightarrow r=\pm 1,2,5,10\\ Possible\ roots: \frac{1}{1},-\frac{1}{1},\frac{1}{2},-\frac{1}{2}, etc.$$
None of the possible roots gives $g(y)=0$. Therefore $g(y)$ is irreducible.
Have I misunderstood the concept of irreducible polynomials?
The polynomial $(x^2+1)^2$ has no roots in $\mathbb{Q}$, but it's reducible.
The root criterion only holds for polynomials of degree two or three, because if they can be factored, at least one of the factors must have degree one (hence a root). On the other hand, a polynomial of degree $>3$ can be the product of two nonconstant polynomials of lower degree having no roots and so be reducible.
Your argument is therefore good for the first polynomial, but not for the second one.
The second polynomial is irreducible over $\mathbb{Q}$ by Eisenstein's criterion applied to the prime $5$, which divides all coefficients except the leading one and its square doesn't divide the constant term.