<s>[INST] 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.
 [/INST] The quadratic formula only gives the roots of the equation, which are unaltered by the constant factor.
Let $\  P(x)=3x^{2}-x-2 \ \text{ and } \ Q(x) = x^2 -\frac{x}{3}-\frac{2}{3}$
The roots of both the polynomials are the same, as 
$$\begin{align}3x^{2}-x-2&=0 \\ \implies x^2 -\frac{x}{3}-\frac{2}{3}&=0\end{align}$$
but $P(x)$ and $Q(x)$ differ by a constant factor $3$, the leading coefficient of $P(x)$.
$$P(x) = 3\times Q(x)$$
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