Question

Partial fractions

2x
------
(x+1)^3

I keep getting 2x = A(x+1)^2 + B(x+1) -2, but I can't solve for A and B. I do know that A+B = 2.

Answer 1

2x/(x+1)^3=A/(x+1)+B/(x+1)^2+ C/(x+1)^3

2x = A(x+1)^2 + B(x+1) +C is correct.
Now set up a system of equations based on degrees of x:
x^2: 0=A
x: 2=2A+B
constants: 0=A+B+C

Solving through substitution you get A=0, B=2, C=-2

So 2x/(x+1)^3=2/(x+1)^2 -2/(x+1)^3

Answer 2

What was that system of equations you did? I'm not sure I understand your method.

Answer 3

@bettirocked

Answer 4

I keep getting 2x = A(x+1)^2 + B(x+1) -2


expand A(x+1)^2 to get Ax^2 +2Ax + A
same for B: Bx+B
we get

0x^2 + 2x + 0 = Ax^2 +2Ax+Bx + A+B-2
0x^2 + 2x + 0 = Ax^2 +(2A+B)x + (A+B-2)

now equate the coefficients of each power of x term:
0 x^2 = A x^2 or 0 = A

2x = (2A+B)x or 2= 2A+B

0 = (A+B-2)

Answer 5

Ohh.....I forgot to add 0x^2 to the left! Thanks Phi!