Note that x^3 -2x^2 - 5x + 6 factors as ( x - 1)(x + 2)(x -3)
Multiplying through by this factorization we get that
x^2 -19 = A( X + 2)(x - 3) + B(x - 1)(x - 3) + C( x -1)(x + 2) simpliify
x^2 -19 = A(x^2 - x - 6) + B(x^2 - 4x + 3) + C(x^2 + x - 2)
1x^2 + 0x - 19 = (A + B + C)x^2 - (A + 4B - C) x + (-6A + 3B - 2C)
Equating coefficients and constant terms we have this system
A + B + C = 1 (1)
-A - 4B + C = 0 (2)
-6A + 3B - 2C = -19 (3)
Add (1) and (2) and we get that
-3B + 2C = 1 (4)
Multiply (2) through by -6
6A + 24B - 6C = 0 add this to (3)
27B -8C = -19 (5)
Multiply (4) by 9
-27B + 18C = 9 add this to (5)
10C = -10
C = -1
And
-3B + 2(-1) = 1
-3B = 3
B = -1
And
A + B + C =1
A - 1 - 1 = 1
A -2 = 1
A = 3
So ({A, B ,C} = { 3, -1, -1 }