`(x + 2)/(x(x^2 - 9))` Write the partial fraction decomposition of the rational expression. Check your result algebraically.

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`(x+2)/(x(x^2-9))`

`(x+2)/(x(x^2-9))=(x+2)/(x(x+3)(x-3))`

Now let `(x+2)/(x(x^2-9))=A/x+B/(x+3)+C/(x-3)`

`(x+2)/(x(x^2-9))=(A(x+3)(x-3)+B(x)(x-3)+C(x)(x+3))/(x(x+3)(x-3))`

`(x+2)/(x(x^2-9))=(A(x^2-9)+B(x^2-3x)+C(x^2+3x))/(x(x^2-9))`

`:.(x+2)=A(x^2-9)+B(x^2-3x)+C(x^2+3x)`

`x+2=Ax^2-9A+Bx^2-3Bx+Cx^2+3Cx`

`x+2=(A+B+C)x^2+(-3B+3C)x-9A`

equating the coefficients of the like terms,

`A+B+C=0`

`-3B+3C=1`

`-9A=2`

Solve the above three equations to get the values of A,B and C,

`A=-2/9`

Bach substitute the value of A in the first equation,

`-2/9+B+C=0`

`B+C=2/9`

From the above equation ,express C in terms of B

`C=2/9-B`

Substitute the expression of C in the second equation,

`-3B+3(2/9-B)=1`

`-3B+2/3-3B=1`

`-6B=1-2/3`

`-6B=1/3`

`B=-1/18`

Now plug the value of B in the expression of C,

`C=2/9-(-1/18)`

`C=2/9+1/18`

`C=(2*2+1)/18`

`C=5/18`

`:.(x+2)/(x(x^2-9))=-2/(9x)-1/(18(x+3))+5/(18(x-3))`

Now let's check the result,

RHS=`-2/(9x)-1/(18(x+3))+5/(18(x-3))`

`=(-2*2(x+3)(x-3)-x(x-3)+5x(x+3))/(18x(x+3)(x-3))`

`=(-4(x^2-9)-(x^2-3x)+5(x^2+3x))/(18x(x^2-9))`

`=(-4x^2+36-x^2+3x+5x^2+15x)/(18x(x^2-9))`

`=(18x+36)/(18x(x^2-9))`

`=(18(x+2))/(18x(x^2-9))`

`=(x+2)/(x(x^2-9))`

=LHS

Hence it is verified.

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