Divide using synthetic division $$ ( 5x^{4}-8x^{3}-5x^{2}+2x+6 ) \div ( x-2 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&5&-8&-5&2&6\\& & 10& 4& -2& \color{black}{0} \\ \hline &\color{blue}{5}&\color{blue}{2}&\color{blue}{-1}&\color{blue}{0}&\color{orangered}{6} \end{array} $$The solution is:
$$ \dfrac{ 5x^{4}-8x^{3}-5x^{2}+2x+6 }{ x-2 } = \color{blue}{5x^{3}+2x^{2}-x} ~+~ \dfrac{ \color{red}{ 6 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&5&-8&-5&2&6\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 5 }&-8&-5&2&6\\& & & & & \\ \hline &\color{orangered}{5}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 5 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&5&-8&-5&2&6\\& & \color{blue}{10} & & & \\ \hline &\color{blue}{5}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 10 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}2&5&\color{orangered}{ -8 }&-5&2&6\\& & \color{orangered}{10} & & & \\ \hline &5&\color{orangered}{2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 2 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&5&-8&-5&2&6\\& & 10& \color{blue}{4} & & \\ \hline &5&\color{blue}{2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 4 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}2&5&-8&\color{orangered}{ -5 }&2&6\\& & 10& \color{orangered}{4} & & \\ \hline &5&2&\color{orangered}{-1}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&5&-8&-5&2&6\\& & 10& 4& \color{blue}{-2} & \\ \hline &5&2&\color{blue}{-1}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}2&5&-8&-5&\color{orangered}{ 2 }&6\\& & 10& 4& \color{orangered}{-2} & \\ \hline &5&2&-1&\color{orangered}{0}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&5&-8&-5&2&6\\& & 10& 4& -2& \color{blue}{0} \\ \hline &5&2&-1&\color{blue}{0}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 0 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}2&5&-8&-5&2&\color{orangered}{ 6 }\\& & 10& 4& -2& \color{orangered}{0} \\ \hline &\color{blue}{5}&\color{blue}{2}&\color{blue}{-1}&\color{blue}{0}&\color{orangered}{6} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{3}+2x^{2}-x } $ with a remainder of $ \color{red}{ 6 } $.