Divide using synthetic division $$ ( 5x^{4}+x^{3}-17x^{2}-3x-12 ) \div ( x-2 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&5&1&-17&-3&-12\\& & 10& 22& 10& \color{black}{14} \\ \hline &\color{blue}{5}&\color{blue}{11}&\color{blue}{5}&\color{blue}{7}&\color{orangered}{2} \end{array} $$The solution is:
$$ \dfrac{ 5x^{4}+x^{3}-17x^{2}-3x-12 }{ x-2 } = \color{blue}{5x^{3}+11x^{2}+5x+7} ~+~ \dfrac{ \color{red}{ 2 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&5&1&-17&-3&-12\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 5 }&1&-17&-3&-12\\& & & & & \\ \hline &\color{orangered}{5}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&5&1&-17&-3&-12\\& & \color{blue}{10} & & & \\ \hline &\color{blue}{5}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 10 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrrr}2&5&\color{orangered}{ 1 }&-17&-3&-12\\& & \color{orangered}{10} & & & \\ \hline &5&\color{orangered}{11}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 11 } = \color{blue}{ 22 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&5&1&-17&-3&-12\\& & 10& \color{blue}{22} & & \\ \hline &5&\color{blue}{11}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -17 } + \color{orangered}{ 22 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}2&5&1&\color{orangered}{ -17 }&-3&-12\\& & 10& \color{orangered}{22} & & \\ \hline &5&11&\color{orangered}{5}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 5 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&5&1&-17&-3&-12\\& & 10& 22& \color{blue}{10} & \\ \hline &5&11&\color{blue}{5}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 10 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrr}2&5&1&-17&\color{orangered}{ -3 }&-12\\& & 10& 22& \color{orangered}{10} & \\ \hline &5&11&5&\color{orangered}{7}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 7 } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&5&1&-17&-3&-12\\& & 10& 22& 10& \color{blue}{14} \\ \hline &5&11&5&\color{blue}{7}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 14 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}2&5&1&-17&-3&\color{orangered}{ -12 }\\& & 10& 22& 10& \color{orangered}{14} \\ \hline &\color{blue}{5}&\color{blue}{11}&\color{blue}{5}&\color{blue}{7}&\color{orangered}{2} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{3}+11x^{2}+5x+7 } $ with a remainder of $ \color{red}{ 2 } $.