`1/(4x^2 - 9)` Write the partial fraction decomposition of the rational expression. Check your result algebraically.

`1/(4x^2-9)=1/[(2x+3)(2x-3)]`

`1/(4x^2-9)=A/(2x+3)+B/(2x-3)`

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Multiply through the the LCD `4x^2-9.`

`1=A(2x-3)+B(2x+3)`

`1=2Ax-3A+2Bx+3B`

`1=(2A+2B)x+(-3A+3B)`

Equate the coefficient of like terms. Then solve for A and B.

`0=2A+2B`

`1=-3A+3B`

Use the elimination method to solve for A and B. Multiply the first equation by 3

and multiply the second equation by 2.

`0=6A+6B`

`2=-6A+6B`

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`2=12B`

`B=1/6`

`0=2A+2B`

`0=2A+2(1/6)`

`A=-1/6`

`A=-1/6, B=1/6`

`1/(4x^2-9)=-1/[6(2x+3)]+1/[6(2x-3)]`

`1/(4x^2-9)=1/[(2x+3)(2x-3)]`

`1/(4x^2-9)=A/(2x+3)+B/(2x-3)`

``

Multiply through the the LCD `4x^2-9.`

`1=A(2x-3)+B(2x+3)`

`1=2Ax-3A+2Bx+3B`

`1=(2A+2B)x+(-3A+3B)`

Equate the coefficient of like terms. Then solve for A and B.

`0=2A+2B`

`1=-3A+3B`

Use the elimination method to solve for A and B. Multiply the first equation by 3

and multiply the second equation by 2.

`0=6A+6B`

`2=-6A+6B`

--------------------

`2=12B`

`B=1/6`

`0=2A+2B`

`0=2A+2(1/6)`

`A=-1/6`

`A=-1/6, B=1/6`

`1/(4x^2-9)=-1/[6(2x+3)]+1/[6(2x-3)]`