The synthetic division table is:
$$ \begin{array}{c|rrr}0&12&21&10\\& & 0& \color{black}{0} \\ \hline &\color{blue}{12}&\color{blue}{21}&\color{orangered}{10} \end{array} $$The solution is:
$$ \frac{ 12x^{2}+21x+10 }{ x } = \color{blue}{12x+21} ~+~ \frac{ \color{red}{ 10 } }{ x } $$Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrr}\color{blue}{0}&12&21&10\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}0&\color{orangered}{ 12 }&21&10\\& & & \\ \hline &\color{orangered}{12}&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 12 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrr}\color{blue}{0}&12&21&10\\& & \color{blue}{0} & \\ \hline &\color{blue}{12}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 21 } + \color{orangered}{ 0 } = \color{orangered}{ 21 } $
$$ \begin{array}{c|rrr}0&12&\color{orangered}{ 21 }&10\\& & \color{orangered}{0} & \\ \hline &12&\color{orangered}{21}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 21 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrr}\color{blue}{0}&12&21&10\\& & 0& \color{blue}{0} \\ \hline &12&\color{blue}{21}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ 0 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrr}0&12&21&\color{orangered}{ 10 }\\& & 0& \color{orangered}{0} \\ \hline &\color{blue}{12}&\color{blue}{21}&\color{orangered}{10} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 12x+21 } $ with a remainder of $ \color{red}{ 10 } $.