Evaluate:

Let

Now putting,

A(x-3)²+B(x+2)(x-3)+C(x+2)=2x+9

Now put x-3=0

Therefore, x=3

A(0)+B(0)+C(3+2) =6+9=15

C=3

Now put x+2=0

Therefore, x=-2

A(-2-3)²+B(0)+C(0) = -4+9=5

Equating the coefficient of x²,we get,

A+B=0

From equation (1), we get,