SOLUTION: Write the partial fraction decomposition of the rational expression: (4x^2+2x-1)/(x^2(x+1))

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Question 637491: Write the partial fraction decomposition of the rational expression:
(4x^2+2x-1)/(x^2(x+1))
Found 2 solutions by Edwin McCravy, DrBeeee:
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
We write the sum of all fractions with a denominator which is a
divisor of the denominator:

%284x%5E2%2B2x-1%29%2F%28x%5E2%28x%2B1%29%29 = A%2Fx + B%2Fx%5E2 + C%2F%28x%2B1%29

Clear of fractions:

4x² + 2x - 1 = Ax(x+1) + B(x+1) + Cx²

Substitute x=-1 to make the (x+1)'s 0

4(-1)² + 2(-1) - 1 = A(-1)(-1+1) + B(-1+1) + C(-1)²

4(1) - 2 - 1 = A(-1)(0) + B(0) + C(1)
   4 - 2 - 1 = 0 + 0 + C
           1 = C

Substitute x=0 to make the 1st and 3rd terms on the right 0:

4x² + 2x - 1 = Ax(x+1) + B(x+1) + Cx²

4(0)² + 2(0) - 1 = A(0)(0+1) + B(0+1) + C(0)

0 + 0 - 1 = A(0)(1) + B(1) + 0
       -1 = 0 + B
       -1 = B

Substitute B = -1 and C = 1

4x² + 2x - 1 = Ax(x+1) + (-1)(x+1) + (1)x²

4x² + 2x - 1 = Ax(x+1) - (x+1) + x²

Substitute x=1  (Note: we could use ANY number here that we haven't used):

4(1)² + 2(1) - 1 = A(1)(1+1) - (1+1) + 1²

 4 + 2 - 1 = 2A - 2 + 1
         5 = 2A - 1
         6 = 2A
         3 = A

So A = 3, B = -1, C = 1

Answer:

%284x%5E2%2B2x-1%29%2F%28x%5E2%28x%2B1%29%29 = 3%2Fx - 1%2Fx%5E2 + 1%2F%28x%2B1%29


Edwin

Answer by DrBeeee(684) About Me  (Show Source):
You can put this solution on YOUR website!
This is tricky because of the x^2 in the denominator you have what is called a double pole at x = 0. This requires two fractions in your expansion, not just one. I suspect that this is the resson you submitted this problem!
Let your given fraction
(4x^2+2x-1)/[x^2*(x+1)] = A/x^2 + B/x + C/(x+1)
You probably had only the A/x^2 term. If so you could never get your original fraction.
The easy way to get A is to multiple both sides of the equation by x^2 and let x=0. You end up with A = constant. Let's do it. Multiply by x^2 yields
(4x^2+2x-1)/(x+1) = A + Bx +Cx^2/(x+1). Now let x=0 and obtain
(-1)/(1) = A + B*0 + C*0/1 or
A = -1 (you should have gotten the same value)
In a similar way we can find C, i.e. multiply both side by (x+1) (the denominator of C) and let x+1 = 0, yields
(4x^2+2x-1)/x^2 = A*(x+1)/x^2 + B*(x+1)/x + C. Now let x+1=0, or x=-1 and get
(4*1+2(-1)-1)/(-1)^2 = A*0 + B*0 + C which reduces to
(4-2-1)/1 = C
C = 1
Now we need to find B. There is a mathematical method involving derivitives etc.. I prefer to use a simpler way using the fact that we have values for A and C. Also remember that the expansion that we are performing must be true for ALL values of x. I use a convenient value such as -1,0,1. Since we used 0 and -1 above let x = 1. The process of getting B is to substitute into the original fraction the values A=-1, C=1 and x=1, which yields
(4+2-1)/(1*2) = -1 + B +1/2 or
5/2 = B - 1/2
B = 5/2 + 1/2
B = 3
Our PFE is
(4x^2+2x-1)/[x^2*(x+1)] = -1/x^2 + 3/x +1/(x+1)
check the expansion by adding the three fraction on the RHS.
RHS = (3x-1)/x^2 + 1/(x+1)
RHS = [(3x-1)(x+1) + x^2]/[x^2*(x+1)]
RHS = [3x^2-x+3x-1+x^2]/[x^2*(x+1)]
RHS = (4x^2+2x-1)/[x^2*(x+1)] = LHS QED