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

##### 1 Answer

#### 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

#(-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

#-6A+B=18#

Subtracting

#19A = -57#

Dividing both sides by

#A = -3#

Putting this value of

#-3+3B=-3#

Hence:

#B = 0#

Putting this value of

#0+3C = 3#

Hance:

#C = 1#

So:

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