$P(x)$ and $Q(x)$

Consider the root of $q(x) = x^2+x+1 = 0$ .

$\begin{aligned} x^2+x+1 & = 0 & \small \color{#3D99F6}{\text{Multiply both sides with }x} \\ x^3+x^2+x & = 0 & \small \color{#3D99F6}{\text{Add 1 on both sides}} \\ x^3+\color{#3D99F6}{x^2+x + 1} & = 1 & \small \color{#3D99F6}{\text{Note that }x^2+x+1=0} \\ x^3 & = 1 \\ \implies x & = \color{#3D99F6}{\omega} & \small \color{#3D99F6}{\text{The third root of unit}} \end{aligned}$

By remainder theorem , we have:

$\begin{aligned} r(x) & = p(\omega) \\ & = 2\omega^{2010} - 5\omega^2 - 13\omega + 7 \\ & = 2 \left(\color{#3D99F6}{\omega^3}\right)^{670} - 5(\color{#D61F06}{\omega^2 + \omega +1}) - 8\omega+ 12 \\ & = 2 \left(\color{#3D99F6}{1}\right)^{670} - 5(\color{#D61F06}{0}) - 8\omega+ 12 \\ & = - 8\omega + 14 \\ \implies r(x) & = 14 - 8x \\ r(2) & = 14 - 8(2) = \boxed{-2} \end{aligned}$

Lets call $x^2 + x +1 = Q(x)$ . So we see by remainder theorem that $r(x)$ cannot be of a degree higher than $1$ . So let $r(x)=ax+b$ .

So we have $p(x) = q(x).f(x) + r(x)$ . Where $f(x)$ is the quotient when $p(x)$ is divided by $q(x)$ .

So $p(x) = (x-\omega)(x-\omega^{2}).f(x) + r(x)$ . Where $\omega$ is a cubic root of unity other than 1.

So $p(\omega) = a\omega + b...........................(1)$ and

$p(\omega^2)=a\omega^{2} + b..............................(2)$ .

we have $(1) -(2)$ gives us $a = -8$ .

And $(1)+(2)$ gives us

$b = 14$ .

Note that in this step we have used the relation $\omega + \omega^{2} + 1 = 0$ .

So we have $r(x) = -8x + 14$ .

So $r(2) = -2$