Question

If `P(x)` is divided by `(x-a)(x-b)` where `a!=b, a,b in RR`, can you prove that the remainder is: `((P(b)-P(a))/(b-a))xx(x-a)+P(a)`?

Answer 1

See below.

Explanation:

`P(x)` can be represented as

`P(x)=Q(x)(x-a)(x-b)+r_1 x+r_2` where `r_1 x+r_2` is the remainder of `(P(x))/((x-a)(x-b))`

The determination of `r_1,r_2` is straightforward

`P(a) = r_1 a+r_2` and

`P(b)=r_1 b + r_2` so, solving for `r_1, r_2` we get

`r_1= (P(b)-P(a))/(b-a)`
`r_2 = -(a P(b)-b P(a))/(b-a)`

and the remainder is

`( (P(b)-P(a))/(b-a) )x - (a P(b)-b P(a))/(b-a) =`

`=( (P(b)-P(a))/(b-a) )(x-a)+P(a)`