How do you express #(5x-1)/(x^2-x-2)# in partial fractions?

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Answer 1 · 2016-11-29T21:46:55

Factorize the denominator, then calculate the numerators of partial fractions.

First you find the zeroes of the denominator to factorize it.

#x^2-x-2 =0#

#x= frac (1+-sqrt(1+8)) 2 = (1+-3)/2#

#x^2-x-2 = (x-2)(x+1)#

Now pose:

#frac (5x-1) ((x-2)(x+1)) = frac A (x-2) + frac B (x+1)#

#frac (5x-1) ((x-2)(x+1)) = frac ( A(x+1)+B(x-2)) ((x-2)(x+1))#

#frac (5x-1) ((x-2)(x+1)) = frac ( Ax+A+Bx-2B) ((x-2)(x+1))#

Equating the coefficient of the same order in the numerators:

#A+B=5#
#A-2B=-1#

and solving the system:

#A=3#
#B=2#

thus:

#frac (5x-1) ((x-2)(x+1)) = frac 3 (x-2) + frac 2 (x+1)#