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: 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.
A: You have to distinguish between “$f_1$ and $f_2$ are the same functions” and
“$f_1$ and $f_2$ have the same roots”.
Clearly $$f_1(x) = a(x^2-Sx+P) = 0 \iff f_2(x) = x^2-Sx+P = 0$$ for any $a \ne 0$. So $f_1$ and $f_2$ have the same roots.
But $f_1 = a \cdot f_2(x) \ne f_2(x)$
for $a \ne 1$. Thus they are not the same functions.
Edit: if you want to get a visual impression: in this Desmos Graph you can modify the value of $a$ (press the play button or modify it manually). Then you see how the function changes but the roots stay the same.
A: $$a(x^2-Sx+P)$$
and 
$$x^2-Sx+P$$
do coincide at the roots (of course, they are both $0$).
The presence of $a$ matters when you evaluate them at other values than the roots !

A: Let $f(x)=a(x^2-Sx+P)$ and $g(x)=x^2-Sx+P$, then these functions have the same roots, (  if $a \ne 0$) but , if $a \ne 1$, these functions are different. For example $f(0)=aP \ne P=g(0)$ (if $P \ne 1$).
A: Two quadratic equations $$ y_1= (x-b)(x-c)$$ and $$y_2= a(x-b)(x-c)$$ have the same roots but not the same values at any other points. They are different function which share common roots.
For example $$ y_1=x^2+5x+6$$ and  $$ y_2=3x^2+15x+18$$ are different functions with the same roots. 
Notice that $$y_1(2) = 20$$ while  $$y_2(2) = 60$$ 
What you like to say is that the two quadratic equations, $$ x^2+5x+6=0$$ and $$ 3x^2+15x+18=0 $$ are equivalent because we can factor $3$ out and $3$ is not zero so it does not change the roots. 
