Find all solutions, other than $2$ for $12x^3-23x^2-3x+2=0$ 
Find all solutions, other than $2$ for $12x^3-23x^2-3x+2=0$  

I started off by taking out an $x$ and got $$x(12x^2-23x-3)+2=0$$
I do not know if this is the correct first step, if it is, then am I able to use the quadratic formula or complete the square to get the answers. Can anyone give me general hints. Please do not solve this for me in anyway. Just give me hints.
 A: HINT
The great bit here is to note that $2$ is a solution. By the factor theorem, we know that the linear factor $(x - 2)$ divides our polynomial.
So perhaps you should divide out $(x-2)$. You'll be left with a quadratic, which we know how to solve very quickly.
A: The preceding answers have provided you with sufficient information. However the following might help you with future endeavors:
How would you factor the polynomial if 2 wasn't given as a root?
There exists a useful theorem: the Rational Root Theorem, that helps
factor polynomials (indirectly). It gurantees that any root of the polynomial will have a numerator that has $c$, the constant, as a factor and a denominator that has $a_n$, the leading coefficient, as a factor.
Therefore by investigating all possible conbinations of the factors of the constant and the factors of the leading coefficient you can eventually arrive at a valid root. The polynomial can then be further factored into a quadratic through synthetic division which can be factored (as you know) through the quadratic formula.
The interesting thing about this method is its large scope of applicability: it not only works on cubic polynomials but on a polynomial with any degree.
A: The reciprocals of the roots $\rm\,r,s,1/2\:$ are roots of the reversed polynomial $\rm\:2\, x^3\! -\! 3\, x^2\! -\! 23\, x\! +\! 12.\: $ Thus by Vieta's Formulas $\rm\:r\!+\!s\!+\!1/2 = 3/2,\ rs/2 = -6,\:$ so $\rm\:r\!+\!s = 1,\ rs = -12,\:$ so $\rm\:r,s = \ldots$
Alternatively, by the Factor Theorem, $\rm\:f(2) = 0\:$ $\Rightarrow$ $\rm\:f(x)\:$ has $\rm\:x\!-\!2\:$ as a factor. Comparing coef's
$$\rm (x-2)(a\, x^2 + b\, x + c)\ =\ 12\, x^3 - 23\,x^2 -3\,x + 2$$
$\rm\qquad x^3\:$ coef $\rm\:\Rightarrow\: a = 12$
$\rm\qquad x^0\:$ coef $\rm\:\Rightarrow\: -2\,c = 2\:\Rightarrow\: c = -1$
$\rm\qquad x^1\:$ coef $\rm\:\Rightarrow\: -3 = c-2b = -1-2b\:\Rightarrow\: b = 1 $
So the quadratic factor is $\rm\: 12\,x^2 + x - 1,\:$ which can be solved by either the Quadratic Formula or Rational Root Test, or $\rm\:(c\,x\!-\!1)\,(d\,x\!+\!1) = 12\,x^2\!+\!x\!-\!1\:$ $\Rightarrow$ $\rm\:cd=12,\ c\!-\!d = 1,\:$ so $\rm\:c,d = \ldots$
