need help again in express the attached equation in partial fraction. the power 3 in the denominator seems to bug me again.
thanks!
(7 +10x -x^2) / (x^3 +3x^2 -x -3)
The denominator,
x^3 +3x^2 -x -3
= (x^3 +3x^2) -(x +3)
= (x^2)(x +3) -(x +3)
= (x +3)(x^2 -1)
= (x +3)(x +1)(x -1)
So,
(7 +10x -x^2) / (x^3 +3x^2 -x -3) = A/(x+3) +B/(x+1) +C/(x-1)
Multiply both sides by (x+3)(x+1)(x-1),
7 +10x -x^2 = A(x+1)(x-1) +B(x+3)(x-1) +C(x+3)(x+1) -----------(i)
When x = -1,
7 +10(-1) -(-1)^2 = 0 +B(-1+3)(-1-1) +0
-4 = B(2)(-2)
B = 1 ---------------**
When x = 1,
7 +10(1) -(1)^2 = 0 +0 +C(1+3)(1+1)
16 = C(4)(2)
C = 2 ---------------**
When x = -3,
7 +10(-3) -(-3)^2 = A(-3+1)(-3-1) +0 +0
7 -30 -9 = A(-2)(-4)
-32 = 8A
A = -32/8 = -4 -------**
Therefore,
(7 +10x -x^2) / (x^3 +3x^2 -x -3) = -4/(x+3) +1/(x+1) +2/(x-1)
Or,
(7 +10x -x^2) / (x^3 +3x^2 -x -3) = 1/(x+1) +2/(x-1) -4/(x+3) -------------answer.