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

1 Answer
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 !