Find the remainder when $x^{100} + 2x + 10$ is divided by $x − 11$ in $\mathbb Z_{17}[x]$ Find the remainder when $x^{100} + 2x + 10$ is divided by $x − 11$ in $\mathbb Z_{17}[x]$
I simplified 
$x^{100} + 2x + 10$ to $x^{15} + 2x + 10$ 
and $x − 11$ to $x+6$ to be in $\mathbb Z_{17}$.
I got stuck here and used Mathematica and got this answer:

Is there any way to get this remainder without actually diving x+6 into $x^{15} + 2x + 10$? 

$\ \ {\rm mod}\ \color{#c00}{17},\, \color{#0a0}{x\!-\!11}\!:\,\ \color{#0a0}{x\equiv 11}\,\Rightarrow\,\color{#0a0}x^{\large \color{#a0f}{100}}\equiv {\color{#0a0}{11}}^{100}\equiv \overbrace{11^4(\color{#c00}{11^{\color{#c00}16}})^6\equiv (-6)^4\color{#c00}1^6}^{\rm little\ \color{#c00}{Fermat}}\equiv 2^2\equiv \color{#a0f}{4}$ 
Therefore we conclude  $\quad \begin{align} &{\color{#0a0}{x}}^{\large\color{#a0f}{100}}+2\color{#0a0}x+10\\ \ \ \equiv \ \ \ &\color{#a0f}4+\,2\,(\color{#0a0}{11})+10\equiv 36\equiv 2\end{align}$
