# 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)?

Feb 24, 2017

See below.

#### Explanation:

$P \left(x\right)$ can be represented as

$P \left(x\right) = Q \left(x\right) \left(x - a\right) \left(x - b\right) + {r}_{1} x + {r}_{2}$ where ${r}_{1} x + {r}_{2}$ is the remainder of $\frac{P \left(x\right)}{\left(x - a\right) \left(x - b\right)}$

The determination of ${r}_{1} , {r}_{2}$ is straightforward

$P \left(a\right) = {r}_{1} a + {r}_{2}$ and

$P \left(b\right) = {r}_{1} b + {r}_{2}$ so, solving for ${r}_{1} , {r}_{2}$ we get

${r}_{1} = \frac{P \left(b\right) - P \left(a\right)}{b - a}$
${r}_{2} = - \frac{a P \left(b\right) - b P \left(a\right)}{b - a}$

and the remainder is

$\left(\frac{P \left(b\right) - P \left(a\right)}{b - a}\right) x - \frac{a P \left(b\right) - b P \left(a\right)}{b - a} =$

$= \left(\frac{P \left(b\right) - P \left(a\right)}{b - a}\right) \left(x - a\right) + P \left(a\right)$