Why is the coefficient of $x$ in $\frac{1}{x}=0$? I usually solve a quadratic equation:
$$ax^2+bx+c=0$$
Through a method I learned in school: For a monic quadratic, you make $x=y-\frac{b}{2}$.
The method is intended for a monic equation but in this case (non-monic equation), I divide all the equation by $a$ to transform it in a monic equation.

Now I'm trying to solve the cubic:
$$x^3+x^2-12x=0$$
For the problem in question, I was thinking if I couldn't do the same, I noticed that the coefficient of $x^2$ is $1$, then I thought about dividing everything by $x^3$ which would give me:
$$\frac{x^2}{x^3}-\frac{12x}{x^3}+1=0\tag{1}$$
$$-12x^{-2}+x^{-1}+1=0\tag{2}$$
$$-\frac{12}{x^2}+\frac{1}{x}+1=0\tag{3}$$
And now, trying to use the same procedure I demonstrated for quadratics, I guess I need to find the coefficient of $x$ in $\frac{1}{x}$ and at this part I'm stuck, If it was the coefficient of $\frac{x}{2}$ (for example), I would just need the coefficient of $x$ divided by $2$ which would be $\frac{1}{2}$, but for $\frac{1}{x}$ I am confused, I tried to use Wolfram|Alpha and it gave me this.
NOTE: I am aware of the existence of the quadratic and cubic formulas, I'm doing this just for curiosity.
 A: Your cubic is already monic, so there's no need to adjust. Furthermore, $$x^3+x^2-12x=x(x^2+x-12)=x(x+4)(x-3)$$ readily, so we can solve your equation without much difficulty.
In answer to your question, though, the reason that the coefficient of $x$ in $1/x$ is $0$ is because there is no term of $x^1$ in the expression $x^{-1}$. The coefficient of $x$ is likewise zero in $x^4+x^2$, in $5x^0$, and in any other combination of powers of $x$ not including $x^1$.
A: ADD With what you've done, you can let $u=1/x$ to move on, however, note that $x=0$ is a solution to the equation, so you can factor $x$ out and move on with a quadratic.

Note that the method you know about quadratic equations has an analog for cubics, however, what you do is anihilate the $x^2$ term. That is, suppose we're given $$x^3+ax^2+bx+c=0$$
Then by letting $x\mapsto y-\frac{a}{3}$ we get $$\eqalign{
  & {\left( {y - \frac{a}{3}} \right)^3} + a{\left( {y - \frac{a}{3}} \right)^2} + b\left( {y - \frac{a}{3}} \right) + c = 0  \cr 
  & {y^3} - 3\frac{a}{3}{y^2} + 3\frac{{y{a^2}}}{9} - \frac{{{a^3}}}{{27}} + a{y^2} - \frac{{2{a^2}}}{3}y + \frac{{{a^3}}}{9} + by - \frac{{ba}}{3} + c = 0  \cr 
  & {y^3} + \left( {b - \frac{{{a^2}}}{3}} \right)y + \left( { - \frac{{ba}}{3} + c - \frac{{{a^3}}}{{27}} + \frac{{{a^3}}}{9}} \right) = 0  \cr 
  & y^3 + Ay + B = 0 \cr} $$ (I hope I made no mistakes up there)
The idea is that we can "depress" any cubic into the form $x^3+ax+b=0$ and there is a formula to solve this equation just as we can solve $x^2+ax+b=0$ using the famous Bhaskara formula. In fact, the first solutions to cubics were for cubics of the form
$$x^3=ax+c$$ $$x^3+ax=c$$
where we make sure the coefficients are kept positive by putting them in the appropriate "side" of $=$.
A: 1/x is not a polynomial in x, so there is no such coefficient. However, 1/x is a rational function in x, with 1/x being formed from the 1/x "monom" with coefficient 1, and that's what wolfram|alpha's Coefficient[] function was thinking about. But it won't help you in your problem.
You have a very simple factor appearing in your equation, which will reduce it to a quadratic problem. 
