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.

$3(x - 1)(x + {2 \over 3}) = (x -1)(3x + 2) = (3x - 3)(x + {2 \over 3}) =...$ etc. are all valid factoring.  The leading coefficient is just a constant.
And if $(x - 1)(x + {2 \over 3}) = 0$ then $3(x - 1)(x + {2 \over 3}) = 0 = (x - 1)(x + {2 \over 3}) $. 
If you are concerned about going from roots to factoring think of it this way:
If the roots of $P(x) = ax^n + ...... + c$ are $r_1, .... ,r_n$ then the polynomial factors as $a(x - r_1)(x - r_2).....(x - r_3) = P(x)$ or $(x - r_1)(x - r_2).....(x - r_3) = P(x)/a$.  When setting to 0 the $a$ doesn't matter as if $y = 0$ then $a*y = 0$ also no matter what the $a$ is.  
Solving for 0, the $a$ gets lost.  Going from the roots back, we only take $(x - r_i)$ so the resulting coefficient will always be $1$.  The resulting polynomial will be $1/a$ of the original.
