expand the quotients to partial fractions?

Question

1. (x+1)/( (x^2 )*(x-1) )
2. x/(x^3-x^2-6x)
3. (x^2+8)/(x^2-5x+6)

Answer 1

#1: Set up the partial fraction decomposition:

(x+1)/(x²(x-1)) = A/x + B/x² + C/(x-1)

Now multiply by x²(x-1):

x+1 = Ax(x-1) + B(x-1) + Cx²

Now, there are two ways to find A, B, and C. The rote method is to expand the polynomial, equate coefficients, and then solve the resulting system of equations. The faster method, which requires a bit of insight, is to make clever substitutions for x that cause most of the variables to cancel out. For instance, substituting x=0:

1 = -B
B=-1

Or x=1:

2 = C

And x=-1:

0 = 2A - 2B + C
0 = 2A + 2 + 2
2A = -4
A=-2

So the partial fraction decomposition is:

-2/x - 1/x² + 2/(x-1)

#2: This problem is greatly simplified if you cancel out the x first:

1/(x²-x-6)

Now factor the denominator:

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

Set up the decomposition:

1/((x-3)(x+2)) = A/(x-3) + B/(x+2)

Multiply by (x-3)(x+2):

1 = (x+2)A + (x-3)B

Substituting x=-2 we find that:

1=-5B
B=-1/5

And x=3:

1=5A
A=1/5

So the decomposition is:

(1/5)/(x-3) - (1/5)/(x+2)

#3: First we factor:

(x²+8)/((x-3)(x-2))

Now, we set up the decomposition, remembering to account for the polynomial part:

(x²+8)/((x-3)(x-2)) = A + B/(x-3) + C/(x-2)

Now, multiply by (x-3)(x-2):

x²+8 = A(x-3)(x-2) + B(x-2) + C(x-3)

Substituting x=2:

12 = -C
C=-12

And x=3:

17 = B

And partially multiplying out the product, we find:

x²+8 = Ax² + stuff, where stuff is entirely composed of x and constant terms.

So equating coefficients of x²:

A=1

Therefore the decomposition is 1 + 17/(x-3) - 12/(x-2).

Answer 2

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

x+1 = Ax(x-1) + B(x-1) +Cx^2
x+1 = Ax^2-Ax +Bx -B +Cx^2 = (A+C)x^2 +(B-A)x -B
So A+C = 0
B-A = 1
-B=1
B= -1
-1-A=1 --> A = -2
-2+C = 0 --> C= 2
so (x+1)/(x^2*(x-1)) = -2/x -1/x^2 +2/(x-1)

In the second problem, you have x/x(x-3)(x+2) = 1/[(x-3)(x+2)]
So 1/[(x-3)(x+2)] = A/(x-3)+B/(x+2)
1/[(x-3)(x+2)] = A(x+2)/[(x-3)(x+2)]+B(x-3)/[(x+2)(x-3)]
So Ax+2A +Bx-3B =1
(A+B)x +2A-3B=1
(A+B)=0
2A-3B)=1
5A = 1
A = .2
B= .8
So x/(x^3-x^2-6x) = .2/(x-3) + .8/(x+2)
You should be able to do the 3rd one now by yourself.

Answer 3

1. [ -1 / x² ] - [ 2 / x ] + [ 2 / x-1 ]
2. [ 1/3 / x+2 ] - [ 1/3 / x-3 ]
3. uhh...

Answer 4

for the second one, why people have different answers?
and if you work backward, it seems none of them are correct.