\[\dfrac{x^2 - 19}{x^3 - 2x^2 -5x + 6} = \dfrac{A}{x-1} + \dfrac{B}{x+2} + \dfrac{C}{x-3}\]
Find constants A, B, and C so that the above expression holds for all real numbers other than 1, -2, and 3.
+129852 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 }
+397 \(\displaystyle \frac{x^{2}-19}{x^{3}-2x^{2}-5x+6} \equiv \frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{x-3} \\ \text{so} \\ x^{2}-19 \equiv A(x+2)(x-3)+B(x-3)(x-1)+C(x-1)(x+2).\)
\(\displaystyle \text{Let }x=1,\qquad -18=A(3)(-2), \qquad A=3. \\ \text{Let }x=-2, \quad \, -15=B(-5)(-3), \qquad B= -1.\\ \text{Let }x=3,\qquad-10=C(2)(5),\qquad C=-1.\)
.
