How do you decompose #(3p-1)/(p^2-1)# into partial fractions?

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Answer 1 · 2016-12-24T14:52:44

#p^2 - 1# can be factored as #(p + 1)(p - 1)#.

#A/(p - 1) + B/(p + 1) = (3p - 1)/((p + 1)(p - 1)#

#A(p + 1) + B(p - 1) = 3p - 1#

#Ap + A + Bp - B = 3P - 1#

#(A + B)p + (A - B) = 3P - 1#

We now write a system of equations.

#{(A + B = 3), (A - B = -1):}#

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

#1 + B = 3#

#B = 2#

Thus, the partial fraction decomposition is #1/(p - 1) + 2/(p + 1)#.

Hopefully this helps!