Math Questions and Answers

Start Your Free Trial

`x/(16x^4 - 1)` Write the partial fraction decomposition of the rational expression. Check your result algebraically.

Expert Answers info

gsarora17 eNotes educator | Certified Educator

calendarEducator since 2015

write762 answers

starTop subjects are Math, Science, and Business

`x/(16x^4-1)`

Let's factorize the denominator,

`16x^4-1=(4x^2)^2-1`

`=(4x^2+1)(4x^2-1)`

`=(4x^2+1)(2x+1)(2x-1)` 

Let `x/(16x^4-1)=A/(2x+1)+B/(2x-1)+(Cx+D)/(4x^2+1)`

`x/(16x^4-1)=(A(2x-1)(4x^2+1)+B(2x+1)(4x^2+1)+(Cx+D)(2x+1)(2x-1))/((2x+1)(2x-1)(4x^2+1))`

`x/(16x^4-1)=(A(8x^3+2x-4x^2-1)+B(8x^3+2x+4x^2+1)+(Cx+D)(4x^2-1))/((2x+1)(2x-1)(4x^2+1))`

`x/(16x^4-1)=(A(8x^3-4x^2+2x-1)+B(8x^3+4x^2+2x+1)+4Cx^3-Cx+4Dx^2-D)/((2x+1)(2x-1)(4x^2+1))`

`x/(16x^4-1)=(x^3(8A+8B+4C)+x^2(-4A+4B+4D)+x(2A+2B-C)-A+B-D)/((2x+1)(2x-1)(4x^2+1))`

`:.x=x^3(8A+8B+4C)+x^2(-4A+4B+4D)+x(2A+2B-C)-A+B-D`

equating the coefficients of the like terms,

`8A+8B+4C=0`        ----- equation 1

`-4A+4B+4D=0`     ----- equation 2

`2A+2B-C=1`            ----- equation 3

`-A+B-D=0`             ------ equation 4

Now we have to solve the above four equations to find the solutions of A,B,C and D.

From equation 1,

`4(2A+2B+C)=0`

`2A+2B+C=0`

Subtract equation 3 from the above equation,

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

`2C=-1`

`C=-1/2`

From equation 2,

`4(-A+B+D)=0`

`-A+B+D=0`

Now subtract equation 4 from the above equation,

`(-A+B+D)-(-A+B-D)=0`

`2D=0`

`D=0`

Now plug in the values of C in the equation 3,

`2A+2B-(-1/2)=1`

`2A+2B+1/2=1`

`2A+2B=1-1/2`

`2(A+B)=1/2`

`A+B=1/4`       ----- equation 5

Plug in the value of D in the equation 4,

`-A+B-0=0`

`-A+B=0`      ---- equation 6

Now add the equations 5 and 6,

`2B=1/4`

`B=1/8`

Plug in the value of B in the equation 6,

`-A+1/8-0`

`A=1/8`

`:.x/(16x^4-1)=(1/8)/(2x+1)+(1/8)/(2x-1)+((-1/2)x)/(4x^2+1)`

`x/(16x^4-1)=1/(8(2x+1))+1/(8(2x-1))-x/(2(4x^2+1))`

 

check Approved by eNotes Editorial