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

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Answer 1 · 2017-02-24T21:36:05

See below.

$P(x)$ 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

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

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

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