Difficulty understanding how $3x^2 + 12$ in $\mathbb{C}[x]$ is reducible.

Question

I am having difficulty understanding how $3x^2 + 12$ in $\mathbb{C}[x]$ is reducible.

It can be factorized into $3(x^2 + 4) = 3[(x + 2i)(x - 2i)]$. However, isn't $3$ a unit in $\mathbb{C}[x]$? After all, real numbers are complex numbers, and $3 * \cfrac{1}{3} = 1$.

Answer 1

You also have $3x^2+12=(3x+6i)(x-2i)$. Neither $3x+6i$ nor $x-2i$ is an unit. So, $3x^2+12$ is reducible.

Answer 2

Yes you are right . But think ,another factorisation is possible for 3x^2+12 in C[x].

3x^2+12 = (3x-6i)(x+2i) So, it is reducible over C.

More over, you can also use the fact

If F is a field ,Any polynomial p of deg 2 is reducible over F if & only if p has a root in F.

Now, since, We know C is algebrically closed field ,hence any polynomial of deg 2 is definitely reducible.