<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] Look at what happens when we complete the square:
$$\begin{array}{lll}
ax^2+bx+c &=& \frac{1}{4a}(4a^2x^2 + 4abx + 4ac)\\
&=& \frac{1}{4a}((2ax)^2 + 2b(2ax) + 4ac)\\
&=& \frac{1}{4a}((2ax)^2 + 2b(2ax) +b^2-b^2+ 4ac)\\
&=& \frac{1}{4a}((2ax+b)^2-(b^2 - 4ac))\\
&=& \frac{1}{4a}((2ax+b)^2-\bigg(\sqrt{b^2 - 4ac}\bigg)^2)\\
&=& \frac{1}{4a}(2ax+b+\sqrt{b^2-4ac})(2ax+b-\sqrt{b^2-4ac})\\
&=& \frac{4a^2}{4a}\bigg(\frac{2ax+b+\sqrt{b^2-4ac}}{2a}\bigg)\bigg(\frac{2ax+b-\sqrt{b^2-4ac}}{2a}\bigg)\\
&=& a\bigg(x-\frac{-b-\sqrt{b^2-4ac}}{2a}\bigg)\bigg(x-\frac{-b+\sqrt{b^2-4ac}}{2a}\bigg)\\
\end{array}$$
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