Divide using synthetic division $$ ( x^{4}-11x^{3}+28x^{2}-15x+6 ) \div ( x-3 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&1&-11&28&-15&6\\& & 3& -24& 12& \color{black}{-9} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{blue}{4}&\color{blue}{-3}&\color{orangered}{-3} \end{array} $$The solution is:
$$ \dfrac{ x^{4}-11x^{3}+28x^{2}-15x+6 }{ x-3 } = \color{blue}{x^{3}-8x^{2}+4x-3} \color{red}{~-~} \dfrac{ \color{red}{ 3 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-11&28&-15&6\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 1 }&-11&28&-15&6\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-11&28&-15&6\\& & \color{blue}{3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 3 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}3&1&\color{orangered}{ -11 }&28&-15&6\\& & \color{orangered}{3} & & & \\ \hline &1&\color{orangered}{-8}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-11&28&-15&6\\& & 3& \color{blue}{-24} & & \\ \hline &1&\color{blue}{-8}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 28 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}3&1&-11&\color{orangered}{ 28 }&-15&6\\& & 3& \color{orangered}{-24} & & \\ \hline &1&-8&\color{orangered}{4}&& \end{array} $$Step 6 : 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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-11&28&-15&6\\& & 3& -24& \color{blue}{12} & \\ \hline &1&-8&\color{blue}{4}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 12 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}3&1&-11&28&\color{orangered}{ -15 }&6\\& & 3& -24& \color{orangered}{12} & \\ \hline &1&-8&4&\color{orangered}{-3}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-11&28&-15&6\\& & 3& -24& 12& \color{blue}{-9} \\ \hline &1&-8&4&\color{blue}{-3}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}3&1&-11&28&-15&\color{orangered}{ 6 }\\& & 3& -24& 12& \color{orangered}{-9} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{blue}{4}&\color{blue}{-3}&\color{orangered}{-3} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-8x^{2}+4x-3 } $ with a remainder of $ \color{red}{ -3 } $.