How do you write the partial fraction decomposition of the rational expression #(6x^2+8x+30)/(x^3-27)#?

1 Answer
Guillaume L.
Apr 8, 2018

#(6x^2+8x+30)/(x^3-27)=4/(x-3)+(2x+2)/(x^2+6x+9)#

Explanation:

#(6x²+8x+30)/(x^3-27)=(6x^2+8x+30)/((x-3)(x^2+3x+9))#
#=A/(x-3)+(Bx+C)/(x^2+3x+9)#
#6x²+8x+30=(A+B)x²+(3A-3B+C)x+(9A-3C)#
Now let's identificate:
#A+B=6#(1)
#3A-3B+C=8#(2)
#9A-3C=30#(3)
#(1)<=>(2)#
So:

#3A-3B+C=8#(1)
#A+B=6#(2)
#9A-3C=30#(3)
#(3)=(3)+3*(1)#
So:
#3A-3B+C=8#(1)
#A+B=6#(2)
#18A-9B=54#(3)
#(3)=((3))/9#
#3A-3B+C=8#(1)
#A+B=6#(2)
#2A-B=6#(3)
#(3)=(3)+(2)#
#3A-3B+C=8#(1)
#A+B=6#(2)
#3A=12#(3)
So:
#A=4#
#3*4-3B+C=8#
#4+B=6#
#A=4#
#B=2#
#12-6+C=8#
So We got #(A,B,C) = (4,2,2)#
So: #(6x^2+8x+30)/(x^3-27)=4/(x-3)+(2x+2)/(x^2+6x+9)#
\0/ here's our answer !

Related questions
Impact of this question
772 views around the world
You can reuse this answer
Creative Commons License