Originally Posted by

**Deveno** Note that we don't actually have to find $p(x)$, which is good (there are too many possibilities).

Here is what we know:

$p(x) = q_1(x)(x - 1) + 4 = q_2(x)(x - 2) + 5$.

We want to find $r(x)$ such that $p(x) = q(x)(x^2 - 3x + 2) + r(x)$.

Note that $r(x) = ax + b$, since $\text{deg}(x^2 - 3x + 2) = 2$.

Let's just play with this a bit.

$r(x) = p(x) - q(x)(x^2 - 3x + 2)$, so:

$r(x) - 4 = p(x) - 4 - q(x)(x^2 - 3x + 2) = q_1(x)(x - 1) + q(x)(x - 2)(x - 1) = [q_1(x) - q(x)(x - 2)](x - 1)$, so $r(x) - 4$ is divisible by $(x - 1)$. This mean that $r(1) - 4 = 0$, that is:

$a + b = 4$.

Similarly,

$r(x) - 5 = p(x) - 5 - q(x)(x^2 - 3x + 2) = q_2(x)(x - 2) + q(x)(x - 2)(x - 1) = [q_2(x) - q(x)(x - 1)](x - 2)$, so $r(x) - 5$ is divisible by $(x - 2)$. What does this say about what $r(2) - 5$ is?