It is just as easy to identify remainders in base $x$ division problems as it is in base $10$ arithmetic.

Look at

$\dfrac {4x{4}-7x{3}+9x{2}-3x-1}{x{2}-x+1}$
on the $1 \leftarrow x$ machine. Can you see that it equals $4x^2-3x+2$ with a remainder of $2x-3$ yet to be divided by $x^2-x+1$?

People typically write this answer as follows:
$\dfrac {4x{4}-7x{3}+9x{2}-3x-1}{x{2}-x+1}=4x2-3x+2+\dfrac{2x-3}{{x{2}-x+1}}$.

Here are some practice problems if you would like to play some more with this idea.

1. Can you deduce what the answer to $\left( 2x^2+7x+7 \right) \div \left( x+2 \right)$ is going to be before doing it?
2. Compute $\dfrac{x4}{x2-3}$.
3. Try this crazy one: $\dfrac{5x5-2x4+x3-x2+7}{x^3-4x+1}$. (If you do it with paper and pencil, you will find yourself trying to draw $84$ dots at some point. Is it swift and easy just to write the number “$84$”? In fact, how about just writing numbers and not bother drawing any dots at all?)

SOLUTIONS

Here are my answers.

1. We know that $\left( 2x^2+7x+6 \right) \div \left( x+2 \right)=2x+3$ so I bet $\left( 2x^2+7x+7 \right) \div \left( x+2 \right)$ turns out to be $2x+3+\dfrac{1}{x+2}$. Does it?
2. $\dfrac {x4}{x2-3}=x2+3+\dfrac{9}{x2-3}$.
3. $5x^2-2x+21+\dfrac {-14x2+82x-14}{x3-4x+1}$.