Is there a rule for spliting the b term of a cubic? I was reading a solved example from a book called "A.Das Gupta" but I did not understand how the author solved the cubic equation. Here's what he did:-
$$2x^3+11x^2+18x-12=0$$
$$2x^3-x^2+12x^2+18x-12=0$$
$$x^2(2x-1)+6(2x^2+3x-2)=0$$
$$x^2(2x-1)+6(2x-1)(x+2)=0$$
$$(2x-1)(x^2+6x+12)=0$$
$$x=\frac{1}{2}$$ 
For the only real root of the polynomial.
I did not understand why the author divided $11x^2$ in $12x^2$ and $-x^2$.
Is there a rule for spliting the $x^2$ term of a cubic?
 A: No, it isn't if you don't know a root beforehand. If I were to factorize that equation, I'll try to check first the roots of the form $p/q$ with $p$ a divisor of $12$ and $q$ a divisor of $2$, the coefficient of $x^3$, and try, as usual, the Ruffini's rule to see wich of these numbers are in fact roots.
Now, if I know that $2x-1$ is a factor, I can try to write the first terms in the form of product of this factor and a "queue", as you see in your example. Split the second term is a good try. Knowing a factor of the polynomial guarantees the result. It's the multiplication algorithm for polynomials in reverse (in the algorithm, the resulting terms from the multiplication of nomomials are grouped)
My idea is that for you to manipulate an equation that way without any other information about it, you had to do a lot of work previously on many other problems.
Try to do it with some polynomilals, e.g. $x-2$ and $2x^2-2x+1$. Multiply them and see how the terms are grouped. Then do the reverse splitting the terms you need to.
