정수와 유리수

There should be $3$ partial products. The second partial product will be the result of multiplying $896$ by $0.$

Notice that the second partial product of all zeros doesn’t really affect the result. We can place a zero as a placeholder in the tens place and then proceed directly to multiplying by the $2$ in the hundreds place, as shown.

Multiply by $10,$ but insert only one zero as a placeholder in the tens place. Multiply by $200,$ putting the $2$ from the $12.$ $2\cdot 6=12$ in the hundreds place.

$$\begin{array}{c} \\ \\ 896\, \\ \underset{\_\_\_\_\_}{\times 201}\, \\ 896\, \\ \underset{\_\_\_\_\_\_\_\_\_\_}{17920}\, \\ 180,096\,\end{array}$$