Divide using synthetic division $$ ( 2x^{3}+12x^{2}-64x-59 ) \div ( 2x-8 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrr}8&2&12&-64&-59\\& & 16& 224& \color{black}{1280} \\ \hline &\color{blue}{2}&\color{blue}{28}&\color{blue}{160}&\color{orangered}{1221} \end{array} $$The solution is:
$$ \dfrac{ 2x^{3}+12x^{2}-64x-59 }{ x-8 } = \color{blue}{2x^{2}+28x+160} ~+~ \dfrac{ \color{red}{ 1221 } }{ x-8 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -8 = 0 $ ( $ x = \color{blue}{ 8 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{8}&2&12&-64&-59\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}8&\color{orangered}{ 2 }&12&-64&-59\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 2 } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{8}&2&12&-64&-59\\& & \color{blue}{16} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 16 } = \color{orangered}{ 28 } $
$$ \begin{array}{c|rrrr}8&2&\color{orangered}{ 12 }&-64&-59\\& & \color{orangered}{16} & & \\ \hline &2&\color{orangered}{28}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 28 } = \color{blue}{ 224 } $.
$$ \begin{array}{c|rrrr}\color{blue}{8}&2&12&-64&-59\\& & 16& \color{blue}{224} & \\ \hline &2&\color{blue}{28}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -64 } + \color{orangered}{ 224 } = \color{orangered}{ 160 } $
$$ \begin{array}{c|rrrr}8&2&12&\color{orangered}{ -64 }&-59\\& & 16& \color{orangered}{224} & \\ \hline &2&28&\color{orangered}{160}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 160 } = \color{blue}{ 1280 } $.
$$ \begin{array}{c|rrrr}\color{blue}{8}&2&12&-64&-59\\& & 16& 224& \color{blue}{1280} \\ \hline &2&28&\color{blue}{160}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -59 } + \color{orangered}{ 1280 } = \color{orangered}{ 1221 } $
$$ \begin{array}{c|rrrr}8&2&12&-64&\color{orangered}{ -59 }\\& & 16& 224& \color{orangered}{1280} \\ \hline &\color{blue}{2}&\color{blue}{28}&\color{blue}{160}&\color{orangered}{1221} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}+28x+160 } $ with a remainder of $ \color{red}{ 1221 } $.