Homework Help Forum: math

Posted by mousey on Saturday, October 24, 2009 at 5:12pm.

using method of partial fractions to decompose this 1/(x-3)(x+3) is 1/6(x-3)-1/6(x+3) correct or have i gone wrong somewhere

• math - jim, Saturday, October 24, 2009 at 7:08pm

  I get 1/(x-3)(x+3) = A/(x-3) + B/(x+3) 1 = (x+3)A + (x-3)B 1 = x(A+B) + 3A -3B -> A + B = 0 -> A = -B -> 3A - 3B = 1 -> A = 1/6; B = -1/6 = (1/6)(x-3) - (1/6)/(x+3)

  
  

• math - mousey, Sunday, October 25, 2009 at 5:45am

  thanks so it is the same thank god for that

  
  

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