Why isn't $y=a(x^2-Sx+P)$ same with $x^2-Sx+P$ If we have roots of the function $y=ax^2+bx+c$ we can calculate $S=\frac{-b}{a}$ and also  $P=\frac{c}{a}$ .
Then we know that we can form the function this way: $$x^2-Sx+P$$
So on the other side we know that we have the function f(x)=y in different ways:
$$y=ax^2+bx+c$$
($\alpha$ and $\beta$ are roots of the quadratic function)
$$y=a(x-\alpha)(x-\beta)$$
And my question is here:
$$y=a(x^2-Sx+P)$$
Actually know that how we can form the qudratic equation using  $x^2-Sx+P$ ,but the function must be like  $y=a(x^2-Sx+P)$.
Actually I don't know that why we add $a$. I know it will be removed when $(a)(\frac{-b}{a})$ 
But I don't know that what is  $y=a(x^2-Sx+P)$ different whitout a!

a is there just to include the possibility of a quadratic having a coefficient other than 1 for $2^{nd}$ degree term $x^2$.  
Otherwise, how will you make a quadratic of the form: $$y=(ax^2+bx+c)$$(where a $\neq$ 1)
 merely by taking it like: $$y=(x-\alpha)(x-\beta)?$$
Clearly, coefficient of $x^2$ is 1 in this.  
So, while assuming a quadratic when its roots are known, we take it $$y=a(x-\alpha)(x-\beta),$$ just to be on the safer side.
