The reason why you are confused is because this thing is really more complicated than it looks. The thing you do is much easier that the reason you do it.
I'll try to go step by step. Do not be discouraged, but the rabbit hole is kinda deep =P
Step 0: You start with the polinomial P
Step 1: Find the roots of the quadratic (as the roots of your equation are ugly, lets use $r_1$ and $r_2$)
Now, if you want to write this polinomial P using only first degree equations (and numbers) and multiplying them, you know that (x-$r_1$) and (x-$r_2$) must appear. (either them or them multplied by a number, really ...)
If (x-$r_1$) does not appear, then the expression you created will not be zero, when you try to evaluate it with $r_1$
If you try to say P=(x+$r_5$)(x-$r_7$), and you put $r_1$ in, you dont get zero.
($r_1$+$r_5$) is not zero and ($r_1$-$r_7$) is not zero either. Multipling two non-zero numbers, we get something that is not zero
(unless you picked $r_5$ as $-r_1$ or $r_7$ as $r_1$ =P)
So we have (x-$r_1$)(x-$r_2$) as an expression that is zero in the right moments. That is nice, but not enough. It still has to match P in all other places.
Step 2: Pick the polinomial P and the polinomial H=(x-$r_1$)(x-$r_2$). Evaluate both on a number that is not a root (say, 1) and compare. If your polinomial H has a result 4 times smaller, that we can say that P=4(x-$r_1$)(x-$r_2$)
- We knew that the polinomial we were trying to factor had (x-$r_1$) and (x-$r_2$) as "factors that involved x". Why only those ? If you add another "factor with x", you are adding another point for the polinomial to be zero. Another root, that it does not have
- That makes it clear why we COULD take that last step: surely there were no further "factors with x". All we could do from there was to multiply by a number
- Other answers will say that the 4 came from the coefficient of $x^2$. That is also true! say you picked H=(x-$r_1$)(x-$r_2$) and did the multiplication: you would get ($x^2$ -x*something + something else). And this should look like the original P, right ? Looking at H, we can tell that, after the multiplication, $x^2$ will appear without a coefficient. So, to get P, we have to multiply H by the coefficient of $x^2$ in P (a.k.a 4)