Partial fraction decomposition

Question

Find the partial fraction decomposition for the rational expressions.  Show the correct decomposition and then solve for the constants. 

A) (x-8)/(x^3 + 4x)

 

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

Answer 1

A> (x-8)/(x(x²+4)= A/x + (Bx+C/(x²+4) 

The numerator on the right is: Ax²+4A + Bx²+Cx

     A+B=0

     4A=-8 => A=-2

     C=-1

then B=2 and finally

(x-8)/(x(x²+4)) = (-2/x) + (2x+1)/(x²+4)

 

B> To do this one you must first divide

 

(x³+2x+1)/(x(x+2) = (x-2) + (6x+1)/(x(x+2) = (x-2) + (A/x) + (B/(x+2))

 

Follow the same procedure and fine A=1/2 and B=5 1/2

Answer 2

 

Denominator D = x(x^2+4)

 

  (x-8)                A              Bx + C

------------ =   -------- + ---------------

x(x^2+4)           x              x^2 + 4

 

x-8 = A(x^2 + 4) + x(Bx+C)

 

x-8 = Ax^2 + 4A + Bx^2 + Cx

 

      = (A+B)x^2 + Cx + 4A

 

A+B=0

C = 1

4A = -8 ----> A = -2 ---->  B=2

 

 

   -2/x + (2x+1)/(x^2+4) =

 

 -2(x^2 + 4) + x(2x+1)/ D

 

(-2x^2 + -8 + 2x^2 + x)/D

 

 = (x-8)/D   ,    so yes it checks

 

----------------------------------------------------

 

For problem B you must divide the polynomials first,

because the numerator has higher degree than the denominator.

 

                   

                __ x____   -4 ______________________

 x^2 + 2x |   x^3 + 0x^2 + 2x   -   1

                   x^3  + 4x^2

                  -------------------------------

                            - 4x^2 + 2x - 1

                           -4x^2  - 8x  

                           ----------------------------

                                      10x - 1

 

So the fraction that needs decomposed is:

 

 10x-1        A            B

--------- = ------ + ------

x(x+2)       x          x+2

 

10x-1 = A(x+2) + Bx

 

10x - 1 = Ax + 2A + Bx

 

10x - 1 = (A+B)x + 2A

 

A+B = 10

2A = -1 ----> A = -1/2

 

B = 10.5

 

-0.5/x + 10.5/(x+2) =

 

which has numerator

[-0.5(x+2) + 10.5x ]=

 

 =0.5X  - 1 + 10.5x = 

 

  10x - 1, so it checks