It matters in a geometric sense.
Take for instance the polynomial $f=x^2 + y^2$. Over $\mathbb{Q}$ (or $\mathbb{R}$), this has a single point of $(0,0)$ as the zero set, but over $\mathbb{C}$, this factors as $f=x^2+y^2=(x-iy)(x+iy)$ giving the zero set a structure as the union of two (complex) lines $x-iy=0$ and $x+iy=0$.
If we know the polynomial $f$ is irreducible over an algebraically closed field $k$, then the zero set of f, $X=Z(f)$ forms a variety (not a union of other polynomial zero sets) with the coordinate ring $A(X)=k[x,y]/(f)$ a domain. You can see quickly that $A(X)$ wont be a domain if $f$ is reducible.