Here, f(x) will be linear function only as highest power in LHS is 4 So, let f(x)=px+q ‌‌
x4
(x2+1)(x−2)
‌(px+q)[x3−2x2+x−2] ‌=‌
+[Ax2+(B−2A)x−2B+cx2+C]
(x2+1)(x−2)
‌⇒‌
x4
(x2+1)(x−2)
‌px4+(q−2p)x3+(p−2q+A+C)x2 ‌=‌
+(q−2p+B−2A)x+(C−2B−2q)
(x2+1)(x−2)
On comparing both sides, we get ‌p=1,q−2p=0⇒q=2 ‌p−2q+A+C=0⇒A+C=3 ‌q−2p+B−2A=0⇒B=2A ‌C−2B−2q=0⇒C−2B=4 From Eqs. (i), (ii) and (iii), we get, A=‌