the last row. Multiply the dropped number by the diagonal before the bar, and place the resulting entries diagonally to the right from the dropped entry. Perform an addition in the next column. Repeat the previous two steps until you would go past the entries at the top with the next diagonal. Then simply add up any remaining columns. Count the terms to the left of the bar. Since there are two, the remainder has degree one and this is the two right-most terms under the bar. Mark the separation with a vertical bar. The terms are written with increasing degree from right to left beginning with degree zero for both the remainder and the result. The result of our division is: For non-monic divisors With a little prodding, the expanded technique may be generalised even further to work for any polynomial, not just monics. The usual way of doing this would be to divide the divisor with its leading coefficient (call it a): then using synthetic division with as the divisor, and then dividing the quotient by a to get the quotient of the original division (the remainder stays the same). But this often produces unsightly fractions which get removed later, and is thus more prone to error. It is possible to do it without first reducing the coefficients of . As can be observed by first performing long division with such a non-monic divisor, the coefficients of are divided by the leading coefficient of after "dropping", and before multiplying. Let's illustrate by performing the following division: A slightly modified table is used: Note the extra row at the bottom. This is used to write values found by dividing the "dropped" values by the leading coefficient of (in this case, indicated by the /3; note that, unlike the rest of the coefficients of , the sign of this number is not changed). Next, the first coefficient of is dropped as usual: and then the dropped value is divided by 3 and placed in the row below: Next, the new (divided) value is used to fill the top rows with multiples of 2 and 1, as in the expanded technique: The 5 is dropped next, with the obligatory adding of the 4 below it, and the answer is divided again: Then the 3 is used to fill the top rows: At this point, if, after getting the third sum, we