# How do you express #(6x^2+8x+30)/(x^3-27)# in partial fractions?

##### 1 Answer

Oct 2, 2016

#### Explanation:

Note that:

#x^3-27 = (x-3)(x^2+3x+9)#

Sticking with Real coefficients, we are looking for a partial fraction decomposition of the form:

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

#color(white)((6x^2+8x+30)/(x^3-27)) = (A(x^2+3x+9)+(Bx+C)(x-3))/(x^3-27)#

#color(white)((6x^2+8x+30)/(x^3-27)) = ((A+B)x^2+(3A-3B+C)x-3C)/(x^3-27)#

So equating coefficients, we get the following system of linear equations:

#{ (A+B=6), (3A-3B+C=8), (-3C=30) :}#

From the third equation we find:

#C = -10#

Substituting this value of

#3A-3B-10=8#

Hence:

#3A-3B=18#

So:

#A-B=6#

Combining this with the first equation, we find:

#A=6#

#B=0#

So:

#(6x^2+8x+30)/(x^3-27) = 6/(x-3) - 10/(x^2+3x+9)#