need help again in express the attached equation inpartial fraction. the power 3 in the denominator seems to bug me again.

http://img266.imageshack.us/img266/5373/math2ga6.gif

thanks! :)

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- October 4th 2007, 12:39 AMsamtrixpartial fraction
need help again in express the attached equation in

**partial fraction**. the power 3 in the denominator seems to bug me again.

http://img266.imageshack.us/img266/5373/math2ga6.gif

thanks! :) - October 4th 2007, 01:11 AMticbol
(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. - October 4th 2007, 07:42 AMsamtrix
thanks ticbol! been a great help! :)