How do you express #(x^4+4x^3+7x^2+8x+6)/(x^3+6x^2+11x+6)# in partial fractions?
1 Answer
Feb 26, 2018
#=x-2+1/(x+1)-1/(x+2)+9/(x+3)#
Explanation:
The coefficients of
#x^3+6x^2+11x+6 = (x+1)(x+2)(x+3)#
So we find:
#(x^4+4x^3+7x^2+8x+6)/(x^3+6x^2+11x+6)#
#=(x^4+6x^3+11x^2+6x-2x^3-12x^2-22x-12+8x^2+24x+18)/(x^3+6x^2+11x+6)#
#=x-2+(8x^2+24x+18)/(x^3+6x^2+11x+6)#
#=x-2+A/(x+1)+B/(x+2)+C/(x+3)#
#=x-2+(A(x+2)(x+3)+B(x+1)(x+3)+C(x+1)(x+2))/(x^3+6x^2+11x+6)#
So:
#8x^2+24x+18 = A(x+2)(x+3)+B(x+1)(x+3)+C(x+1)(x+2)#
Putting
#2 = 8-24+18 = 2A" "# so#A=1#
Putting
#2 = 32-48+18 = -B" "# so#B=-1#
Putting
#18 = 72-72+18 = 2C" "# so#C=9#
So:
#(x^4+4x^3+7x^2+8x+6)/(x^3+6x^2+11x+6)#
#=x-2+1/(x+1)-1/(x+2)+9/(x+3)#