a)
We use partial fractions to find A, B and C such that:
10x2+101x+89(x+9)2(x−1)=Ax+9+B(x+9)2+Cx−1
Dividing both sides of the equation by 1(x+9)2(x−1) we obtain:
A(x+9)(x−1)+B(x−1)+C(x+9)2=10x2+101x+89
⇒(A+C)x2+(8A+B+18C)x+−9A−B+81C=10x2+101x+89
Identifying coefficients:
Coefficient x2: A+C=10
Coefficent x: 8A+B+18C=101
Constant term: −9A−B+81C=89
On solving these equations we obtain A=8, B=1 and C=2
Which gives:10x2+101x+89(x+9)2(x−1)=8x+9+1(x+9)2+2x−1
Apply same method to solve b) and c)