# Question #60e34

Oct 1, 2016

$r \left(x\right) = - 2 x + 5$

#### Explanation:

Calling

$P \left(x\right) = {\left(x - 3\right)}^{200} - {\left(x - 2\right)}^{100}$ and
$p \left(x\right) = {x}^{2} - 5 x + 6 = \left(x - 2\right) \left(x - 3\right)$

we have

$P \left(x\right) = p \left(x\right) Q \left(x\right) + r \left(x\right)$ in which

$Q \left(x\right)$ is the quocient polynomial and
$r \left(x\right) = a x + b$ is the remainder

Then we have

$P \left(2\right) = 1 = 2 a + b$ and
$P \left(3\right) = - 1 = 3 a + b$

Solving for $a , b$ we have $a = - 2 , b = 5$ so

$r \left(x\right) = - 2 x + 5$