Let $A$, $B$, and $C$ be constants so that 1x3−x2−21x+45=Ax+5+Bx−3+C(x−3)2holds for all real numbers $x$ other than $-5$ and $3.$ What is $A$?
x^3 - x^2 -21x + 45 factors as (x-3)^2 (x + 5)
Multiply through by this factorization and we get
1 = A(x-3)^2 + B (x-3) ( x + 5) + C(x + 5)
Let x = -5
Then A(-8)^2 = 1 → A = 1/64
+28 
