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,






1