Use long division to divide the first polynomial below by the second.

Question

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$$equire{enclose}
\begin{array}{c|cccccc}
\multicolumn{2}{r}{11ebfdc1_ec23_abe3} &  &  &  &  &    \
\cline{2-7}
11ebfdc1_ec23_b307_abe3_2f8f26dab6c4_TBX8727_11 & 1\ \ & 1\ \ & e\ \ & b\ \ & f\ \ & d\ \ c_1\ c_2\ c_3 \
\enclose{longdiv}{11ebfdc1_ec23_da1a_abe3} &  &  &  &  &  &  \
\multicolumn{2}{r}{11ebfdc1_ec23_2890} &  &  &  &  &  \
\cline{2-5}
\multicolumn{2}{r}{0\ \ \ 0\ \ e\ \ b} & f\ \ & d\ \ & c_1\ c_2\ c_3 & & \
\multicolumn{2}{r}{} & & \multicolumn{1}{|@{}c}{0} & & & \
\multicolumn{2}{r}{} & &  & d\ \ & 7\ c_2\ c_3 & \
\multicolumn{2}{r}{} & & & & \multicolumn{1}{|@{}c}{0} & \
\multicolumn{2}{r}{} & & & &  & 9\ c_2\ c_3 \
\end{array}
$$

The divisor is $11ebfdc1\_ec23\_b307\_abe3\_2f8f26dab6c4\_TBX8727\_11$, and the dividend is $11ebfdc1\_ec23\_abe3\_c_1c_2c_3dfbe$. We start by dividing the first two digits of the dividend by the first digit of the divisor to get $1$, which is the first digit of the quotient. This is also our partial product, so we subtract $(11ebfdc1\_ec23\_b307\_abe3\_2f8f26dab6c4\_TBX8727\_11) \cdot 1$ from the dividend to get $11ebfdc1\_ec23\_2890\_c_1c_2c_3dfbe$.

Next, we divide $eb$ by $1b$ to get $e$, which is the second digit of the quotient. Our next partial product is $(11ebfdc1\_ec23\_b307\_abe3\_2f8f26dab6c4\_TBX8727\_11) \cdot e = 11ebfdc1\_ec23\_8be7\_736c$, and we subtract this from the previous result to get $0\_0\_ebc_1c_2c_3dfbe$.

We continue this process, bringing down the next digit and dividing it by the divisor until we have our final quotient: $1\ 1\ e\ b\ d\ 9\ c_2\ c_3$. Therefore, the correct answer is C.

Use Long Division to Divide the First Polynomial Below by the Second