How do you write #(3x)/((x + 2)(x - 1))# as a partial fraction decomposition?

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Answer 1 · 2016-08-04T13:28:42

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

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

Getting every term over common denominator yields:

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

When I do these I eliminate one bracket by setting x to a particular value, but there are many other methods.

#color(red)(x = 1: ) #

#3 = 3B implies B = 1#

#color(red)(x=-2: )#

#-6 = -3A implies A = 2#

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