A certain polynomial `P(x)x in R` when divided by k`x-a ,x-ba n dx-c` leaves remainders`a , b ,a n dc` , resepectively. Then find remainder when `P(x)` is divided by `(x-a)(x-b)(x-c)w h e r eab, c` are distinct.

Explanation

By Remainder theorem `f(a)=a,f(b)=b` and `f(c)=c`
Let the quotient be `Q(x)` and remainder is `R(x)`.
`:.f(x)=(x-a)(x-b)(x-c)Q(x)+R(x)`
`:.f(a)=0+R(a)impliesR(a)=a`
`f(b)=0+R(b)impliesR(b)=b` and `f(c)=0+R(c)`
`impliesR(c)=c` ltbr. So the equation `R(x)-x=0` has there roots a,b and c. But its degree is atmost two. So `R(x)-x` must be zero polynomial (or identity)
Hence `R(x)=x`.