A Missing Term

When a power of x is missing, write it with a 0 in front before you divide. In 407 the 0 does the same job for the tens. So ${x}^{3}-8$ becomes ${x}^{3}+0{x}^{2}+0x-8$. Now each subtraction has the same powers of x lined up.

You stop when what is left has a lower power of x than the number you divide by. For $x-2$, you stop at a plain number. For a divisor with ${x}^{2}$, you can stop at $3x+1$.