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

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Answer 1 · 2016-11-24T14:51:51

the partial fraction decomposition is #2/(x + 2) + 1/(x- 1)#

We can factor the denominator as #(x + 2)(x- 1)#.

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

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

#Ax - A + Bx + 2B = 3x#

#(A + B)x + (2B - A) = 3x#

We can now write a system of equations.

#{(A + B = 3), (2B - A = 0):}#

Solving:

#B = 3 - A#

#2(3 - A) - A = 0#

#6 - 2A - A = 0#

#-3A = -6#

#A = 2#

#A + B = 3#

#2 + B = 3#

#B = 1#

Hence, the partial fraction decomposition is #2/(x + 2) + 1/(x- 1)#.

Hopefully this helps!