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# Decompose the following expression into a sum of partial fractions `18/(x^2-81)`  ...

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Salutatorian

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Decompose the following expression into a sum of partial fractions

`18/(x^2-81)`

18/`x^(2)` -81 (partial fraction decompostion)

Posted by sara212 on July 15, 2013 at 7:28 AM via web and tagged with calculus, decomposition, factorize, math, partial fractions, precalculus, systems of equations

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`18/(x^2-81)` .

First factor the denominator (which is a difference of two squares):

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

As we have two unknowns, it is customary to assign A and B for the unknown values and to split the denominator:

`therefore 18/((x-9)(x+9))= A/(x-9) + B/(x+9)`  Now multiply out the denominator:

`therefore (18(x-9)(x+9))/((x-9)(x+9))`   `= (A(x-9)(x+9))/(x-9) + (B(x-9)(x+9))/(x+9)`

Cross-cancel:

`therefore 18= A(x+9) +B(x-9)`

Remove the brackets and then group the x-es and the constants:

`therefore 18= Ax+9A + Bx -9B = Ax+Bx+9A-9B`

`therefore 18= x(A+B) + 9(A-B)` Apportion a value of x=0 to enable simplification:

`therefore 18= 0(A+B)+9(A-B)`

`therefore 18=9(A-B)` As we still have 2 unknowns apportion another value to x to allow simplification. So reverting to

`18=x(A+B) +9(A-B)` use x=9:

`therefore 18= 9(A+B) +9(A-B)` Expand:

`therefore 18=9A +9B +9A - 9B` Simplify as the Bs cancel out:

`therefore 18= 18A `

`therefore A=1`

To find B use the previous equation from when x=0 which was reduced to :

`18= 9(A-B)` Now substitute the value for A:

`therefore 18=9(1-B)`

`therefore 18-9 = -B`

`therefore B=-9`

Now you have the numerators :

Ans: `therefore 18/(x^2-81) = 1/(x-9) - 9/ (x+9)`

Posted by durbanville on July 15, 2013 at 8:04 AM (Answer #1)