Partial fractions question?

Question

decompose the following partial fraction:
(-4x^3 - 3x^2 + 12) / (x^4 + 2x^3)

the denominator should reduce to x^3 (x + 2) which gives

original fraction = A / x + B / x^2 + C / x^3 + D / (x + 2)

then: -4x^3-3x^2+12 = A(x^2)(x+2) + B (x)(x+2) + C(x+2) + D(x^3)

I can isolate and solve for C and D by plugging in x=-2 and x=0 but how can I isolate A and B??

Answer 1

( - 4x³ - 3x² + 12)/ (x⁴ + 2x³) = A/x + B/x² + C/x³ + D/(x + 2) -------------------->(i)

Rewrite the expression with common denominator. and Cancel common denominator terms.

- 4x³ - 3x² + 12 = A(x²)(x + 2) + B (x)(x + 2) + C(x + 2) + D(x³)

- 4x³ - 3x² + 12 = A(x³ + 2x²) + B (x² + 2x) + C(x + 2) + D(x³)

- 4x³ - 3x² + 12 = Ax³ + 2Ax² + Bx² + 2Bx + Cx + 2C + D(x³)

- 4x³ - 3x² + 12 = x³[A + D] + x²[2A + B] + x(2B + C) + 2C

Compare the coefficients x³, x², x and constant.

x³ coefficients is A + D = - 4-------------------->(1)

x² coefficients is 2A + B = - 3----------------->(2)

x coefficients is 2B + C = 0-------------------->(3)

Constant is 2C = 12

Divide each side by 2. then C = 6

Substitute C = 6 in the equation (3).

Then 2B + 6 = 0

Subtract 6 from each side.

2B = - 6

Divide each side by 2. then B = - 3.

Substitute B = -3 in the equation (2).

2A - 3 = - 3

Add 3 to each side. then 2A = 0

Divide each side by 2. then A = 0.

Substitute A = 0 in the equation (1). Then D = - 4

Substitute A = 0, B = -3. C = 6 and D = -4 in the equation (i).

 - 4x³ - 3x² + 12 = A(x²)(x + 2) + B (x)(x + 2) + C(x + 2) + D(x³).