Partial Fractions need help?

Question

question:

f(x) = 2x^2 / (2x+1)(x+1)

Given that f(x) = A + B /(2x+1) + C / (x +1)

Find the values for A, B and C

THanks

Answer 1

This is the same as when you
Multiply B/(2x+1) times (x+1)/(x+1)
Multiply C/(x+1) times (2x+1) / (2x +1)
Multiply A times (2x+1)(x+1) / (2x+1)(x+1)
To get a common denominator
the same as in the first fraction

A(2x+1)(x +1) + B(x +1) + C(2x +1)
-----------------------------------------------
(2x +1)(x +1)

So
A(2x+1)(x +1) + B(x +1) + C(2x +1)
= 2x^2
the original numerator

To find constant C plug in x = -1
0 + 0 + C(-2 +1) = 2
C = -2
To find constant B plug in x = -1/2
A + B(1/2) + 0 = 1/2
B = 1
To find constant A plug in x = 0
A + B + C = 0
Since C = -2 and B = 1 then
A + 1 - 2 = 0
A = 1

So A = 1, B = 1, C = -2
f(x) = 1 + 1/(2x+1) - 2/(x+1)

Source:
http://tutorial.math.lamar.edu/AllBrowsers/2414/PartialFractions.asp

Answer 2

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

Let x = -1 so that x +1 = 0.
Step 1.
In the LHS leave the factor (x +1) and write the remaining as 2x^2 / (2x+1) and find the value by putting x = -1.
We get 2x^2 / (2x+1) = -2.

In the right hand side take the fraction which contains (x+1) alone i.e Cand equate it to the above value. C = -2.
.

Step 2.

Similarly put x = - ½ so that (2x+1) = 0. Leaving it we get

2x^2 / (x+1) and this is equal to 1 when x= -1/2.

So B= 1.

Step 3.

Comparing the coefficients of x^2 in LHS and RHS , 2A = 2 or
Letting x =0 we get

A +B+C = 0 so A = 1.

Thus
2x^2 / (2x+1)(x+1) = 1 + 1 /(2x+1) - 2 / (x +1)
-----------------------------------------------------------------------------

OR

RHS = {A (2x+1) (x +1) + B(x +1) + C(2x+1) } / (2x+1)(x+1).

Hence

2x^2 = {A (2x+1) (x +1) + B(x +1) + C(2x+1) }

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

Comparing the coefficients of like terms ,

2A = A

(3A +B+ 2C) = 0 and
(A +B+C ) = 0

Solving we get , A= 1 , B=1 and C = -2.

Answer 3

let 2x^2/(2x+1)(x+1)=A/(2x+1)+B/(x+1)+C
2x^2=A(x+1)+B(2x+1).+C(x+1)(2x+1)
put x=-1,2=-B, B=-2
put x=0, 0=A+B+C,
put x=-1/2,1/2=A/2,A=1,C=-1=2=1
Pf's are 2/(2x+1)-2/(x+1).+1
verifyRHS=2[1/(2x+1)]-1/(x+1)+1/2]