How do you express #(-5s-36) / [ (s+2) (s^2+9) ]# in partial fractions?

Question

Sadie Alexander · Accepted Answer

The answer is #=-2/(s+2)+(2s-9)/(s^2+9)# Let's start the decomposition in partial fractions #(-5s-36)/((s+2)(s^2+9))=A/(s+2)+(Bs+C)/(s^2+9)# #=(A(s^2+9)+(Bs+C)(s+2))/((s+2)(s^2+9))# #:.-5s-36=A(s^2+9)+(Bs+C)(s+2)# let #s=-2##=>##-26=13A##=>###A=-2 #-36=9A+2C##=>##C=-9# coeficients of s, #-5=2B+C##=>##B=2# #:.(-5s-36)/((s+2)(s^2+9))=-2/(s+2)+(2s-9)/(s^2+9)#

Anna Allen · Answer

undefined To express (-5s-36) / [ (s+2) (s^2+9) ] in partial fractions, follow these steps:

1. Factor the denominator completely.
   (s^2+9) factors as (s+3i)(s-3i) where i is the imaginary unit.

2. Write the given expression as the sum of partial fractions with undetermined coefficients.
   (-5s-36) / [ (s+2) (s^2+9) ] = A/(s+2) + (Bs+C)/(s^2+9)

3. Multiply both sides by the common denominator to clear the fractions.
   -5s-36 = A(s^2+9) + (Bs+C)(s+2)

4. Expand and equate coefficients of like terms.
   -5s-36 = As^2 + 9A + Bs^2 + 2Bs + Cs + 2C

5. Group terms with the same powers of s.
   -5s-36 = (A+B)s^2 + (2B+C)s + (9A+2C)

6. Equate coefficients:
   For s^2: A + B = 0
   For s: 2B + C = -5
   Constant term: 9A + 2C = -36

7. Solve the system of equations for A, B, and C.

8. Once A, B, and C are found, substitute them back into the partial fraction decomposition.