The synthetic division table is:
$$ \begin{array}{c|rrr}3&4&-10&9\\& & 12& \color{black}{6} \\ \hline &\color{blue}{4}&\color{blue}{2}&\color{orangered}{15} \end{array} $$The solution is:
$$ \frac{ 4x^{2}-10x+9 }{ x-3 } = \color{blue}{4x+2} ~+~ \frac{ \color{red}{ 15 } }{ x-3 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrr}\color{blue}{3}&4&-10&9\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}3&\color{orangered}{ 4 }&-10&9\\& & & \\ \hline &\color{orangered}{4}&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 4 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrr}\color{blue}{3}&4&-10&9\\& & \color{blue}{12} & \\ \hline &\color{blue}{4}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 12 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrr}3&4&\color{orangered}{ -10 }&9\\& & \color{orangered}{12} & \\ \hline &4&\color{orangered}{2}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 2 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrr}\color{blue}{3}&4&-10&9\\& & 12& \color{blue}{6} \\ \hline &4&\color{blue}{2}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ 6 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrr}3&4&-10&\color{orangered}{ 9 }\\& & 12& \color{orangered}{6} \\ \hline &\color{blue}{4}&\color{blue}{2}&\color{orangered}{15} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x+2 } $ with a remainder of $ \color{red}{ 15 } $.