`x^2/(x^4 - 2x^2 - 8)` Write the partial fraction decomposition of the rational expression. Check your result algebraically.

`x^2/(x^4-2x^2-8)`

Let's factorize the denominator,

`x^4-2x^2-8=x^4-4x^2+2x^2-8`

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

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

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

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

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

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

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

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

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

Now,

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

Equating the coefficients of like terms,

`A+B+C=0`     --- equation 1

`-2A+2B+D=1`   ------ equation 2

`2A+2B-4C=0`     ------ equation 3

`-4A+4B-4D=0`   ----- equation 4

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

Rewrite equation 1 ,

`A+B=-C`

Substitute above in equation 3,

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

`2(-C)-4C=0`

`-2C-4C=0`

`-6C=0`

`C=0`

Rewrite equation...

`x^2/(x^4-2x^2-8)`

Let's factorize the denominator,

`x^4-2x^2-8=x^4-4x^2+2x^2-8`

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

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

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

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

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

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

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

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

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

Now,

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

Equating the coefficients of like terms,

`A+B+C=0`     --- equation 1

`-2A+2B+D=1`   ------ equation 2

`2A+2B-4C=0`     ------ equation 3

`-4A+4B-4D=0`   ----- equation 4

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

Rewrite equation 1 ,

`A+B=-C`

Substitute above in equation 3,

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

`2(-C)-4C=0`

`-2C-4C=0`

`-6C=0`

`C=0`

Rewrite equation 2 as,

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

Substitute the above in equation 4,

`-4A+4B-4D=0`  

`2(-2A+2B)-4D=0`

`2(1-D)-4D=0`

`2-2D-4D=0`

`2-6D=0`

`D=2/6`

`D=1/3`

Plug the values of C in equation 3 ,

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

`2A+2B=0`

`2(A+B)=0`

`A+B=0`    ------ equation 5

Plug the value of D in equation 4,

`-4A+4B-4(1/3)=0`

`-4A+4B=4/3`

`4(-A+B)=4/3`

`-A+B=1/3`        ---- equation 6

Solve equations 5 and 6 to find the solutions of A and B,

Add the equations 5 and 6.

`2B=1/3`  

`B=1/6`

Plug the value of B in equation 5,

`A+1/6=0`

`A=-1/6`

`:.x^2/(x^4-2x^2-8)=(-1/6)/(x+2)+(1/6)/(x-2)+(0*x+1/3)/(x^2+2)`

`x^2/(x^4-2x^2-8)=-1/(6(x+2))+1/(6(x-2))+1/(3(x^2+2))`