The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&1&-5&0&1\\& & -2& 14& \color{black}{-28} \\ \hline &\color{blue}{1}&\color{blue}{-7}&\color{blue}{14}&\color{orangered}{-27} \end{array} $$The solution is:
$$ \frac{ x^{3}-5x^{2}+1 }{ x+2 } = \color{blue}{x^{2}-7x+14} \color{red}{~-~} \frac{ \color{red}{ 27 } }{ x+2 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&-5&0&1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 1 }&-5&0&1\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 1 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&-5&0&1\\& & \color{blue}{-2} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrr}-2&1&\color{orangered}{ -5 }&0&1\\& & \color{orangered}{-2} & & \\ \hline &1&\color{orangered}{-7}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&-5&0&1\\& & -2& \color{blue}{14} & \\ \hline &1&\color{blue}{-7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 14 } = \color{orangered}{ 14 } $
$$ \begin{array}{c|rrrr}-2&1&-5&\color{orangered}{ 0 }&1\\& & -2& \color{orangered}{14} & \\ \hline &1&-7&\color{orangered}{14}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 14 } = \color{blue}{ -28 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&-5&0&1\\& & -2& 14& \color{blue}{-28} \\ \hline &1&-7&\color{blue}{14}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -28 \right) } = \color{orangered}{ -27 } $
$$ \begin{array}{c|rrrr}-2&1&-5&0&\color{orangered}{ 1 }\\& & -2& 14& \color{orangered}{-28} \\ \hline &\color{blue}{1}&\color{blue}{-7}&\color{blue}{14}&\color{orangered}{-27} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-7x+14 } $ with a remainder of $ \color{red}{ -27 } $.