The synthetic division table is:
$$ \begin{array}{c|rrrr}2&4&-3&5&-6\\& & 8& 10& \color{black}{30} \\ \hline &\color{blue}{4}&\color{blue}{5}&\color{blue}{15}&\color{orangered}{24} \end{array} $$The remainder when $ 4x^{3}-3x^{2}+5x-6 $ is divided by $ x-2 $ is $ \, \color{red}{ 24 } $.
We can find remainder using synthetic division method.
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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&4&-3&5&-6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 4 }&-3&5&-6\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 4 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&4&-3&5&-6\\& & \color{blue}{8} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 8 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}2&4&\color{orangered}{ -3 }&5&-6\\& & \color{orangered}{8} & & \\ \hline &4&\color{orangered}{5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 5 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&4&-3&5&-6\\& & 8& \color{blue}{10} & \\ \hline &4&\color{blue}{5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 10 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrr}2&4&-3&\color{orangered}{ 5 }&-6\\& & 8& \color{orangered}{10} & \\ \hline &4&5&\color{orangered}{15}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 15 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&4&-3&5&-6\\& & 8& 10& \color{blue}{30} \\ \hline &4&5&\color{blue}{15}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 30 } = \color{orangered}{ 24 } $
$$ \begin{array}{c|rrrr}2&4&-3&5&\color{orangered}{ -6 }\\& & 8& 10& \color{orangered}{30} \\ \hline &\color{blue}{4}&\color{blue}{5}&\color{blue}{15}&\color{orangered}{24} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 24 }\right) $.