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

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

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Answer 1 · 2016-11-05T09:49:30

The answer is #=-2/(s+2)+(2s-9)/(s^2+9)#

Explanation:

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)#

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

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