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