In each case a speci64257c example is used to illustrate the general principle this is not meant as a exhaustive presentation of the method Recall that to use partial fractions the degree of the polynomial in the numerator must be strictly less than ID: 44185
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2Coveringup(s�1)2intheLHSandevaluatingats=1givesA=2.MovingtheAtermtotheLHSandsimplifyinggives3s2�16s+21 (s�1)2(s+3)�2 (s�1)2=3s2�16s+21�2(s+3) (s�1)2(s+3)=3s2�18s+15 (s�1)2(s+3)so3s2�18s+15 (s�1)2(s+3)=B s�1+C s+3:Itmustnowbethecasethat(s�1)isafactorofboththenumeratorandthedenominatorontheLHS.Thiscanbeseensincetheright-hand-side(RHS)isthepartialfractionexpansionofarationalfunctionwith(s�1)(s+3)inthedenominator.Inanyevent,dividingthenumeratorby(s�1)leaves3s�15sowehave3s�15 (s�1)(s+3)=B s�1+C s+3:NownolinearfactorsinthedenominatorarerepeatedsowecanquicklyndB=�3andC=6.Thus3s2�16s+21 (s�1)2(s+3)=2 (s�1)2�3 s�1+6 s+3:3.LinearandquadraticfactorsindenominatorThiscanbecombinedwiththeabovecasestohandlearangeofdi erentcases.g(s) (s2+ps+q)(s�s0)=As+B s2+ps+q+C s�s0FindCrstbycover-upmethodwiths=s0thenmoveCtermtotheLHSandsimplify:g(s)�C(s2+ps+q) (s2+ps+q)(s�s0)=As+B s2+ps+qYouwillndthatg(s)�C(s2+ps+q)hasafactorof(s�s0).Afterdividingthisfactoroutyouwillobtaing(s) (s2+ps+q)=As+B s2+ps+qandAandBcanbedeterminedbyinspection.Notethatiftherearemultiplequadraticfactorsthismethodwilllikelynotbeofmuchhelp.Example:3s (s2+2)(s�1)=As+B s2+2+C s�1Bycoveringup(s�1)intheLHSandsubstitutings=1wendthatC=1.NowmovetheCtermtotheLHSandsimplify3s (s2+2)(s�1)�1 s�1=3s�s2�2 (s2+2)(s�1)=�(s�1)(s�2) (s2+2)(s�1)=2�s s2+2so2�s s2+2=As+B s2+2whichgivesA=�1andB=2.Thus3s (s2+2)(s�1)=2�s s2+2+1 s�1: