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,