<s>[INST] How can I factor out $x^2+ax-2 (3a+1)(2a+1)$? I have tried it this way but could not factor it out :

$$x^2+ax-2 (3a+1)(2a+1)$$

Here, let
     $$3a+1=m\tag{i}$$
$$2a+1=n\tag{ii}$$
So, (i)-(ii) gives us $$m-n=a$$ Then 
$$\begin{align}
x^2+ax-2(3a+1)(2a+1) &= x^2+(m-n)x-2mn \\
&= x^2+mx-nx-mn-mn \\
&= x(x+m)-n(x+m)-mn \\
&= (x+m)(x-n)-mn\end{align}$$
And then I got stuck.Please help!
Thanks in advance.
 [/INST] Writing out the two equations
$$
x^2+ax-2 (3a+1)(2a+1)=(x-b)(x-c)=x^2-(b+c)x+bc
$$
and comparing coefficients gives the equations
$$
a+b+c=0,\quad  12a^2 +10a +bc + 2=0.
$$
Substituting $c=-b-a$ and solving for $b$ we obtain
$$
b=\frac{\pm \sqrt{49a^2 + 40a + 8} - a}{2}
$$
Edit: Following the suggestion of lab bhattacharjee, cancelling the factor $2$, we obtain the much nicer solutions
$$
b=2a+1, \text{ or } b=-3a-1
$$
</s>