Question

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

Answer 1

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

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 `C` into the second equation, we get:

`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)`