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^4+4x^3+7x^2+8x+6)/(x^3+6x^2+11x+6)#

#=x-2+1/(x+1)-1/(x+2)+9/(x+3)#

Explanation:

The coefficients of #x^3+6x^2+11x+6# may already be familiar to you, since it factors as:

#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 #x=-1# we find:

#2 = 8-24+18 = 2A" "# so #A=1#

Putting #x=-2# we find:

#2 = 32-48+18 = -B" "# so #B=-1#

Putting #x=-3# we find:

#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)#