Question 637488
Given:
(1) [x^2+12x+12]/[x^3-4x]
First step is to factor the denominator of (1)
(2) (x^3-4x) = x(x^2-4) = x(x-2)(x+2)
Our expansion of (1) is
(2) [x^2+12x+12]/[x(x-2)(x+2)] = A/x + B/(x-2) + C/(x+2)
The PFE technique is used to find the values of the constants A, B, and C. To get each of the constants, multiply both sides of (2) by the denominator of that constant, then set that denominator equal to zero.
To get A, multiply both sides of (2) by x, then let x equal to zero.
Multiplying (2) by x yields
(3) [x^2+12x+12]/[(x-2)(x+2)] = A + Bx/(x-2) + Cx/(x+2)
Now let x = 0 in (3) to get
(4) [12]/[(-2)(2)] = A 
A = -3
To get B, multiply both sides of (2) by (x-2) and let (x-2) = 0, or x=2.
(5) [2^2+12*2+12]/[2(2+2)] = B
B = 5
To get C, multiply both sides of (2) by (x+2) and let (x+2) = o, or x=(-2).
(6) [(-2)^2+12(-2)+12]/[(-2)(-2-2)] = C
C = -1
Let's see if these values of A, B, and C are correct.
Is (1) = (2)?
Is {[x^2+12x+12]/(x^3-4x) = -3/x + 5/(x-2) - 1/(x+2)}?
Is {[ditto] = [-3(x-2)(x+2) + 5x(x+2) - x(x-2)]/[x(x-2)(x+2)]}?
Is {[ditto] = [-3(x^2-4) +5x^2 +10x -x^2 +2x]/x(x^2-4)]}?
Is {[ditto] = [-3x^2 +12 +5x^2 +10x -x^2 +2x]/[x^3-4x]}?
Is {[x^2+12x+12]/[x^3-4x] = [x^2+x12x+12]/[x^3-4x]}? Yes
Answer: 9x^2+12x+12)/(x^3-4x) = -3/x + 5/(x-2) - 1/(x+2)