<s>[INST] How to factor quadratic $ax^2+bx+c$? How do I shorten this? How do I have to think?
$$ x^2 + x - 2$$
The answer is 
 $$(x+2)(x-1)$$
I don't know how to get to the answer systematically. Could someone explain?
Does anyone have a link to a site that teaches basic stuff like this? My book does not explain anything and I have no teacher; this is self-studies.
Please help me out; thanks!
 [/INST] As Ross pointed out, and as was previously discussed, we know that
$$
(x+a)(x+b) = x(x+b) + a(x+b) = x^2 + bx + ax + ab = x^2 + (a+b)x + ab.
$$
Therefore, to factor a quadratic expression $x^2 + cx + d$, all one has to do is find two numbers that multiply to $d$ and add to $c$.  Let $m$ and $n$ be those two numbers that add to $c$ and multiply $d$.  Then:
$$
x^2 + cx + d = (x+m)(x+n).
$$
Note that it doesn't matter which order we pick the two numbers since $a+b = b+a$ and $ab = ba$.
In your case, to factor $x^2 + x - 2 = x^2 + 1x - 2$, we need to find two numbers that add to $1$ and multiply to $-2$.  Just by picking and trying different numbers, we find the two numbers are $2$ and $-1$.  
A good way to go about finding the numbers is to determine all the factors of the constant term, which for us is $-2$.  Since $-2 = 2 \cdot (-1) = (-2) \cdot 1$, we only have to check whether $2 + (-1)$ or $(-2) + 1$ equals one.  Since it is the first pair of numbers with this property, these are the two numbers we were looking for.  Therefore,
$$
x^2 + x - 2 = (x+2)(x+(-1)) = (x+2)(x-1).
$$

The reason this works:  Suppose that we want to factor $x^2 + cx + d$, and we have found the numbers $m$ and $n$ such that $m + n = c$ and $mn = d$.  Then:
$$
\begin{align}
x^2 + cx + d &= x^2 + (m+n)x + mn
\\
&= x^2 + mx + nx + mn
\\
&= x(x+m) + n(x+m)
\\
&= (x+n)(x+m)
\end{align}
$$
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