The synthetic division table is:
$$ \begin{array}{c|rrrr}0&1&5&1&2\\& & 0& 0& \color{black}{0} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{1}&\color{orangered}{2} \end{array} $$The solution is:
$$ \frac{ x^{3}+5x^{2}+x+2 }{ x } = \color{blue}{x^{2}+5x+1} ~+~ \frac{ \color{red}{ 2 } }{ x } $$Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrr}\color{blue}{0}&1&5&1&2\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}0&\color{orangered}{ 1 }&5&1&2\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 1 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&1&5&1&2\\& & \color{blue}{0} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 0 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}0&1&\color{orangered}{ 5 }&1&2\\& & \color{orangered}{0} & & \\ \hline &1&\color{orangered}{5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 5 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&1&5&1&2\\& & 0& \color{blue}{0} & \\ \hline &1&\color{blue}{5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 0 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrr}0&1&5&\color{orangered}{ 1 }&2\\& & 0& \color{orangered}{0} & \\ \hline &1&5&\color{orangered}{1}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 1 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&1&5&1&2\\& & 0& 0& \color{blue}{0} \\ \hline &1&5&\color{blue}{1}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 0 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}0&1&5&1&\color{orangered}{ 2 }\\& & 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{1}&\color{orangered}{2} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+5x+1 } $ with a remainder of $ \color{red}{ 2 } $.