How do you express #(4x^3) / (x^3 + 2x^2 - x - 2)# in partial fractions?
1 Answer
Nov 25, 2017
Explanation:
Note that:
#x^3+2x^2-x-2 = (x^3+2x^2)-(x+2)#
#color(white)(x^3+2x^2-x-2) = x^2(x+2)-1(x+2)#
#color(white)(x^3+2x^2-x-2) = (x^2-1)(x+2)#
#color(white)(x^3+2x^2-x-2) = (x-1)(x+1)(x+2)#
So:
#(4x^3)/(x^3+2x^2-x-2) = 4+A/(x-1)+B/(x+1)+C/(x+2)#
So:
#4x^3 = 4(x^3+2x^2-x-2)+A(x+1)(x+2)+B(x-1)(x+2)+C(x-1)(x+1)#
Putting
#4 = 6A" "# so#" "A=2/3#
Putting
#-4 = -2B" "# so#" "B=2#
Putting
#-32 = 3C" "# so#" "C = -32/3#
So:
#(4x^3)/(x^3+2x^2-x-2) = 4+2/(3(x-1))+2/(x+1)-32/(3(x+2)#