Factor 3x^2+12x+5

I just want to 'play' with this question for a moment.

I am going to cosider the equation

0=3x^2+12x+5

$${\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{12}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{21}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}\right)}{{\mathtt{3}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{21}}}}{\mathtt{\,-\,}}{\mathtt{6}}\right)}{{\mathtt{3}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{3.527\: \!525\: \!231\: \!651\: \!946\: \!7}}\\
{\mathtt{x}} = -{\mathtt{0.472\: \!474\: \!768\: \!348\: \!053\: \!3}}\\
\end{array} \right\}$$

So I could say

$$3x^2+12x+5 = \left(x+\frac{\sqrt{21}+6}{3}\right)$$ $$\left(x-\frac{\sqrt{21}-6}{3}\right)$$ and I could factor out 9.

I would do that but LaTex is playing up and I cannot be bothers with it anymore!

Anyway, I can do that can't I?

OK...let's see if it's "factorable.'

(3x +1) (x + 5)    Nope !!

(3x +5) (x +1)     Nope !!

These are the only possibilities we have, because 3 and 5 are prime.

Thus, the expression isn't "factorable."