Evaluate:
Let
Now putting,
A(x2+4)+(Bx+C)(x+2)= 8
Putting x+2=0,
X=-2
A(4+4)+0=8
A=1
By equating the coefficient of x2 and constant term, A+B=0
1+B=0
B=-1
4A+2C=8
4×1+2C=8
2C=4
C=2
From equation (1),we get,