How do you express # (3x^2 - 4x - 2) / [(x-1)(x-2)]# in partial fractions?
1 Answer
Aug 7, 2016
Explanation:
#(3x^2-4x-2)/((x-1)(x-2))#
#=(3x^2-4x-2)/(x^2-3x+2)#
#=(3x^2-9x+6+5x-8)/(x^2-3x+2)#
#=(3(x^2-3x+2)+5x-8)/(x^2-3x+2)#
#=3 + (5x-8)/(x^2-3x+2)#
#=3 + (5x-8)/((x-1)(x-2))#
#=3 + A/(x-1) + B/(x-2)#
Use Heaviside's cover-up method to find:
#A=(5(1)-8)/((1)-2) = (-3)/(-1) = 3#
#B=(5(2)-8)/((2)-1) = 2/1 = 2#
So:
#(3x^2-4x-2)/((x-1)(x-2)) = 3 + 3/(x-1) + 2/(x-2)#
