So I need to find $A$, $B$ and $C$ where: $A(x-1)(x+3) + B(x-1) + C$ is equivalent to $2x^2 + 12x + 7$. I know it has something to do with using alpha and beta but I don't know how to approach it. Help is very much appreciated.
(Thank you this Question has been Solved)
Since$$A(x+1)(x+3)+B(x+1)+C=Ax^2+(4A+B)x+3A+B+C,$$it's just a matter of solving the system$$\left\{\begin{array}{l}A=2\\4A+B=12\\3A+B+C=7.\end{array}\right.$$I don't know what are those $\alpha$ and $\beta$ that you mentioned.
A slightly simpler method could be to substitute in $x=1$ first, then $x=-3$.
For $2x^2 + 12 x + 7 = A (x-1)(x+3) + B(x-1) + C$ to hold for all $x$, it looks convenient to try $x=1$. Then, $2+12+7 = C$, so $C=21$.
Trying $x=-3$, $-4B+C = 18-24+7$, so $-4B+21=1$, and hence $B=5$.
To find $A$, we can instead look at the $x^2$ coefficients, so $A=2$.
$y = A(x-1)(x+3) + B(x-1) + C =2x^2 + 12x + 7\\ y(1) = C = 2+12+ 7 = 21\\ y(-3) = -4B + C = 18-36+7 = -11$
Using $C = 21$, we find $B = 8$
$A$ is the coefficient of the $x^2$ term.
$y = 2(x-1)(x+3) + 8(x-1) + 21 =2x^2 + 12x + 7$
