The synthetic division table is:
$$ \begin{array}{c|rrr}2&4&15&10\\& & 8& \color{black}{46} \\ \hline &\color{blue}{4}&\color{blue}{23}&\color{orangered}{56} \end{array} $$The solution is:
$$ \frac{ 4x^{2}+15x+10 }{ x-2 } = \color{blue}{4x+23} ~+~ \frac{ \color{red}{ 56 } }{ x-2 } $$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.
$$ \begin{array}{c|rrr}\color{blue}{2}&4&15&10\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}2&\color{orangered}{ 4 }&15&10\\& & & \\ \hline &\color{orangered}{4}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrr}\color{blue}{2}&4&15&10\\& & \color{blue}{8} & \\ \hline &\color{blue}{4}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ 8 } = \color{orangered}{ 23 } $
$$ \begin{array}{c|rrr}2&4&\color{orangered}{ 15 }&10\\& & \color{orangered}{8} & \\ \hline &4&\color{orangered}{23}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 23 } = \color{blue}{ 46 } $.
$$ \begin{array}{c|rrr}\color{blue}{2}&4&15&10\\& & 8& \color{blue}{46} \\ \hline &4&\color{blue}{23}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ 46 } = \color{orangered}{ 56 } $
$$ \begin{array}{c|rrr}2&4&15&\color{orangered}{ 10 }\\& & 8& \color{orangered}{46} \\ \hline &\color{blue}{4}&\color{blue}{23}&\color{orangered}{56} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x+23 } $ with a remainder of $ \color{red}{ 56 } $.