The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&-3&0&0&-4&5\\& & -6& -12& -24& \color{black}{-56} \\ \hline &\color{blue}{-3}&\color{blue}{-6}&\color{blue}{-12}&\color{blue}{-28}&\color{orangered}{-51} \end{array} $$The solution is:
$$ \frac{ -3x^{4}-4x+5 }{ x-2 } = \color{blue}{-3x^{3}-6x^{2}-12x-28} \color{red}{~-~} \frac{ \color{red}{ 51 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&-3&0&0&-4&5\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ -3 }&0&0&-4&5\\& & & & & \\ \hline &\color{orangered}{-3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&-3&0&0&-4&5\\& & \color{blue}{-6} & & & \\ \hline &\color{blue}{-3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}2&-3&\color{orangered}{ 0 }&0&-4&5\\& & \color{orangered}{-6} & & & \\ \hline &-3&\color{orangered}{-6}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&-3&0&0&-4&5\\& & -6& \color{blue}{-12} & & \\ \hline &-3&\color{blue}{-6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrrr}2&-3&0&\color{orangered}{ 0 }&-4&5\\& & -6& \color{orangered}{-12} & & \\ \hline &-3&-6&\color{orangered}{-12}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&-3&0&0&-4&5\\& & -6& -12& \color{blue}{-24} & \\ \hline &-3&-6&\color{blue}{-12}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ -28 } $
$$ \begin{array}{c|rrrrr}2&-3&0&0&\color{orangered}{ -4 }&5\\& & -6& -12& \color{orangered}{-24} & \\ \hline &-3&-6&-12&\color{orangered}{-28}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -28 \right) } = \color{blue}{ -56 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&-3&0&0&-4&5\\& & -6& -12& -24& \color{blue}{-56} \\ \hline &-3&-6&-12&\color{blue}{-28}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -56 \right) } = \color{orangered}{ -51 } $
$$ \begin{array}{c|rrrrr}2&-3&0&0&-4&\color{orangered}{ 5 }\\& & -6& -12& -24& \color{orangered}{-56} \\ \hline &\color{blue}{-3}&\color{blue}{-6}&\color{blue}{-12}&\color{blue}{-28}&\color{orangered}{-51} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -3x^{3}-6x^{2}-12x-28 } $ with a remainder of $ \color{red}{ -51 } $.