Find at least two ways to find $a, b$ and $c$ in the parabola equation I've been fighting with this problem for some hours now, and i decided to ask the clever people on this website. 
The parabola with the equation $y=ax^2+bx+c$ goes through the points $P, Q$ and $R$. How can I find $a, b$ and $c$ in at least two different ways, when
$P(0,1), Q(1,0)$ and $R(-1,3)$?

Let $y=A(x-\alpha)(x-\beta)+B(x-\beta)(x-\gamma)+C(x-\gamma)(x-\alpha)$ where $\alpha,\beta,\gamma$ are the abscissa of $P,Q,R$ 
So,  $y=Ax(x-1)+B(x-1)(x+1)+C(x+1)x$ putting $\alpha=0,\beta=1,\gamma=-1$
As the parabola passes through $P(0,1)$ so it will satisfy the equation of the parabola 
$\implies 1=B(0-1)(0+1)\implies B=-1$
Similarly, for $Q(1,0), 0=C(1+1)\implies C=0$ and  for $R(-1,3), 3=A(-1)(-1-1)\implies A=\frac32$
So, $y=\frac32x(x-1)+(-1)(x-1)(x+1)+0(x+1)x=\frac{x^2}2-\frac32x+1$
But $y=ax^2+bx+c$
Now compare with the coefficients of the different powers of $x$ 
