About Factorization I have some issues understanding factorization.
If I have the expression $x^{2}-x-7$ then (I was told like this) I can put this expression equal to zero and then find the solutions with the quadratic formula, so it gives me $x_{0,1}= 1 \pm 2\sqrt{2}$ then $$x^{2}-x-7 = (x-1-2\sqrt{2})(x-1+2\sqrt{2}).$$
That is correct I have checked it.
Now for the expression $3x^{2}-x-2$ if I do the same I have $x_{0} = 1$ and $x_1=\frac{-2}{3}$ so I would have $$3x^{2}-x-2 = (x-1)(x+\frac{2}{3})$$
but this is not correct since  $(x-1)(x+\frac{2}{3}) = \frac{1}{3}(3x^{2}-x-2)$,
the correct factorization is $3x^{2}-x-2 = (3x+2)(x-1)$.
So I guess finding the roots of a quadratic expression is not sufficient for factorizing.

For equations of the form $ax^2 + bx + c$ you can use the quadratic equation $x = \frac{-b \pm \sqrt{b^2-4ac }}{2a}$ to help you factor, like you were doing. The adjustment you need to make is to multiply your factorization by a. 
In the example you had, $3x^2 -x - 2$, you found the roots were x = 1 and x $= \frac{-2}{3}$. If you do $3*(x+\frac{2}{3})(x−1)$, then you have (3x+2)(x−1) when you multiply. This works in general. I hope this helps.
