Question

How do you write the partial fraction decomposition of the rational expression `(-3x^2 +3x - 5) / (3x^3 +x^2 + 6x +2)`?

Answer 1

`(-3x^2+3x-5)/(3x^3+x^2+6x+2) = -3/(3x+1)+1/(x^2+2)`

Explanation:

Given:

`(-3x^2+3x-5)/(3x^3+x^2+6x+2)`

Note that the denominator factors as:

`3x^3+x^2+6x+2 = (3x+1)(x^2+2)`

The quadratic factor `(x^2+2)` is irreducible over the reals, so we are (probably) looking for a partial fraction decomposition of the form:

`(-3x^2+3x-5)/(3x^3+x^2+6x+2) = A/(3x+1)+(Bx+C)/(x^2+2)`

`color(white)((-3x^2+3x-5)/(3x^3+x^2+6x+2)) = (A(x^2+2)+(Bx+C)(3x+1))/(3x^3+x^2+6x+2)`

`color(white)((-3x^2+3x-5)/(3x^3+x^2+6x+2)) = ((A+3B)x^2+(B+3C)x+(2A+C))/(3x^3+x^2+6x+2)`

Equating coefficients, this gives us a system of equations:

`{ (A+3B=-3), (B+3C=3), (2A+C=-5) :}`

Subtracting `3` times the third equation from the second, we get:

`-6A+B=18`

Subtracting `3` times this equation from the first equation, we get:

`19A = -57`

Dividing both sides by `19`, we find:

`A = -3`

Putting this value of `A` into the first equation, we get:

`-3+3B=-3`

Hence:

`B = 0`

Putting this value of `B` into the second equation, we get:

`0+3C = 3`

Hance:

`C = 1`

So:

`(-3x^2+3x-5)/(3x^3+x^2+6x+2) = -3/(3x+1)+1/(x^2+2)`