Is $6 | 2a(3b+3)$ true $\forall a,b \in \Bbb Z$?

Question

I have a homework question regarding unique factorisation, asking for me to find whether the following is true:

$6\;\mid\;2a(3b+3)\;\forall \text{ }a,b \in \Bbb Z$

I've tried the following:

$6\mid2a(3b + 3) $

$ =6\mid6ab + 6\color{red}a $

Since $a$ and $b$ are integers, the product of $6ab$ is divisible by 6, therefore $6\;\mid\;2a(3b+3)$ is true for all $a,b \in \Bbb Z$

Does this constitute an adequate proof or should I approach this another way?

Answer 1

Notice that $$2a(3b+3) = 6ab+6\color{red}a.$$ Now you can simply write $$2a(3b+3) = \color{orange}6 \cdot (\color{blue}{ab+a}) + \color{green}0.$$