How do you express #(2x-2)/((x-5)(x-3))# in partial fractions?

1 Answer
Feb 24, 2016

#(2x-2)/((x-5)(x-3))" "=" " 4/(x-5)- 2/(x-3)#

Explanation:

Write as:#" " (2x-2)/((x-5)(x-3))" "=" " A/(x-5)+B/(x-3)#

Thus: #" " (2x-2)/((x-5)(x-3)) =(A(x-3)+B(x-5))/((x-5)(x-3)) #

So:#" "2x-2" "=" "A(x-3)+B(x-5)#

#2x-2" "=" "Ax-3A+Bx-5B#

Collecting like terms

#2x-2" "=" "(A+B)x -3A-5B#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Comparing LHS to RHS

#2x=(A+B)x" so "A+B =2" "#............................(1)

#-2=-3A-5B" so "B=(2-3A)/5" "#...................(2)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Substitute (2) into (1) giving:

#A+(2-3A)/5=2#

#(5A+2-3A)/5=2#

#2A=(2xx5)-2 =8#

#A=4" "#.......................................(3)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Substitute (3) into (1) giving:

#4+B=2 #

#B=-2#
'~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)((2x-2)/((x-5)(x-3))" "=" " 4/(x-5)- 2/(x-3))#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check: #4(x-3)-2(x-5) = 4x-12-2x+10 = 2x-2#

Matching original numerator so ok!