Friday, January 10, 2014

`x^3/((x + 2)^2(x - 2)^2)` Write the partial fraction decomposition of the rational expression.

`x^3/((x+2)^2(x-2)^2)`


Let `x^3/((x+2)^2(x-2)^2)=A/(x+2)+B/(x+2)^2+C/(x-2)+D/(x-2)^2`


`x^3/((x+2)^2(x-2)^2)=(A(x+2)(x-2)^2+B(x-2)^2+C(x-2)(x+2)^2+D(x+2)^2)/((x+2)^2(x-2)^2)`


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


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


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


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


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


equating the coefficients of the like terms,


`A+C=1`


`-2A+B+2C+D=0`


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


`8A+4B-8C+4D=0`


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


Rewrite the third equation as,


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


substitute the expression of ( A+C) from the first equation in the above equation,


`-4(1)-4B+4D=0`


`-4B+4D=4`


`4(-B+D)=4`


`-B+D=1`


`D=1+B`


Express C in terms of A from first equation,


`C=1-A`  


Substitute the expressions...

`x^3/((x+2)^2(x-2)^2)`


Let `x^3/((x+2)^2(x-2)^2)=A/(x+2)+B/(x+2)^2+C/(x-2)+D/(x-2)^2`


`x^3/((x+2)^2(x-2)^2)=(A(x+2)(x-2)^2+B(x-2)^2+C(x-2)(x+2)^2+D(x+2)^2)/((x+2)^2(x-2)^2)`


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


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


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


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


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


equating the coefficients of the like terms,


`A+C=1`


`-2A+B+2C+D=0`


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


`8A+4B-8C+4D=0`


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


Rewrite the third equation as,


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


substitute the expression of ( A+C) from the first equation in the above equation,


`-4(1)-4B+4D=0`


`-4B+4D=4`


`4(-B+D)=4`


`-B+D=1`


`D=1+B`


Express C in terms of A from first equation,


`C=1-A`  


Substitute the expressions of C and D in the second equation,


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


`-2A+B+2-2A+1+B=0`


`-4A+2B=-3` (equation 5)


Substitute the expressions of C and D in the fourth equation,


`8A+4B-8(1-A)+4(1+B)=0`


`8A+4B-8+8A+4+4B=0`


`16A+8B=4`


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


`4A+2B=1` (equation 6)


Solve equation 5 and 6 to get the solutions of A and B,


Add equation 5 and 6,


`4B=-3+1`


`4B=-2`


`B=-2/4`


`B=-1/2`


Plug the value of B in the equation 5,


`-4A+2(-1/2)=-3`


`-4A-1=-3`


`-4A=-3+1`


`-4A=-2`


`A=1/2`


Now plug the values of A and B in the expressions of C and D,


`C=1-1/2`


`C=1/2`


`D=1+(-1/2)`


`D=1/2`


`:.x^3/((x+2)^2(x-2)^2)=1/(2(x+2))-1/(2(x+2)^2)+1/(2(x-2))+1/(2(x-2)^2)`



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