How do you express #1/ [ (x-1)(x+2)(x-3)] # in partial fractions?

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Answer 1 · 2018-06-25T20:30:48

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#1/( (x-1)(x+2)(x-3) ) #

# = A/(x-1) + B/(x+2) + C/(x-3) #

#=> 1 = A(x+2)(x-3) +B(x-1)(x-3) +C(x-1)(x+2) #

Let #r = -2 # Solving to give #B= 1/15 #

Let #r = 3# Solving to give #C =1/10 #

Let #r = 1 # Solving to give #A = -1/6 #

Hence our new partial fraction form is...

# = - 1/(6(x-1)) + 1/(15(x+2)) + 1/(10(x-3)) #