x+P(x−1)(x−3)=Qx−1+2x−3
x+P(x−1)(x−3)=Q(x−3)+2(x−1)(x−1)(x−3)
Comparing LHS and RHS of the equation, we have
x+P=Qx−3Q+2x−2
P=−3Q−2
Q+2=1impliesQ=1−2=−1
P=−3(−1)−2=3−2=1
P+Q=1+(−1)=0