+1348 How do I solve this?
Let $A$, $B$, and $C$ be constants so that
\[\frac{x^2-21x-3}{x^3-x^2-21x+45}=\frac{A}{x+5}+\frac{B}{x-3} + \frac{C}{(x - 3)^2}\]
holds for all real numbers $x$ other than $-5$ and $3.$ What is $A$?
+129771 Note that x^3 - x^2 -21x + 45 = (x + 5) (x - 3) (x - 3)^2
Multiply through by this factorization and we have
x^2 - 21x -3 = A(x - 3)^2 + B( x + 5) ( x-3) + C( x + 5)
x^2 -21x - 3 = A( x^2 - 6x + 9) + B( x^2 + 2x -15) + C(x + 5)
Equate coefficients on each side
1 = A + B → B = 1 - A (1)
-21 = -6A + 2B + C (2)
-3 = 9A -15B + 5C (3)
Sub (1) into (2) , (3)
-21 = -6A + 2 ( 1 - A) + C → -21 = - 6A + 2 - 2A + C → -8A + C = -23 (4)
-3 = 9A - 15 ( 1 - A) + 5C → -3 = 9A - 15 + 15A + 5C → 24A + 5C = 12 (5)
Multiply (4) through by (-5) add to (5)
40A - 5C = 115
24A + 5C = 12
____________
64 A = 127
A = 127/64
