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)#?

1 Answer
Feb 24, 2017

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)#