Show that $2x^2 + 4$ is reducible in $\Bbb C$ Show that $2x^2 + 4$ is reducible in $\Bbb C$.  

By the definition of irreducible, $2x^2 + 4$ is reducible if I can find $f(x)$ and $g(x)$ such that $2x^2 + 4 = f(x)g(x)$ and $f(x)$ and $g(x)$ are both not units.  But I can't seem to find a non-unit product of $2x^2 + 4$ in $\Bbb C$.
A polynomial of degree $2$ over a field (like $\mathbb{C}$ or $\mathbb{R}$ or $\mathbb{Q}$) is reducible exactly when it has a root.

In your case the polynomial does have a root in $\mathbb{C}$ so it is reducible.

$$
2x^2 + 4 = 2(x- i\sqrt{2})(x + i\sqrt{2})
$$

In fact the Fundamental Theorem of Algebra states that all polynomials over $\mathbb{C}$ are reducible if their degree is greater than $1$.

If you consider the polynomial over $\mathbb{R}$ (or $\mathbb{Q}$), then the polynomial is irreducible because it does not have any roots.
polynomial over C[x] does not have a unit when x is put to be roots...
As 2(x-i√2)(x+i√2) for x to be +-i√2 factors are not unit over C[x]