Divide using synthetic division $$ ( 5x^{3}+10x^{2}-x-18 ) \div ( -x-2 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&5&10&-1&-18\\& & 10& 40& \color{black}{78} \\ \hline &\color{blue}{5}&\color{blue}{20}&\color{blue}{39}&\color{orangered}{60} \end{array} $$The solution is:
$$ \dfrac{ 5x^{3}+10x^{2}-x-18 }{ x-2 } = \color{blue}{5x^{2}+20x+39} ~+~ \dfrac{ \color{red}{ 60 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&5&10&-1&-18\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 5 }&10&-1&-18\\& & & & \\ \hline &\color{orangered}{5}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&5&10&-1&-18\\& & \color{blue}{10} & & \\ \hline &\color{blue}{5}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ 10 } = \color{orangered}{ 20 } $
$$ \begin{array}{c|rrrr}2&5&\color{orangered}{ 10 }&-1&-18\\& & \color{orangered}{10} & & \\ \hline &5&\color{orangered}{20}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 20 } = \color{blue}{ 40 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&5&10&-1&-18\\& & 10& \color{blue}{40} & \\ \hline &5&\color{blue}{20}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 40 } = \color{orangered}{ 39 } $
$$ \begin{array}{c|rrrr}2&5&10&\color{orangered}{ -1 }&-18\\& & 10& \color{orangered}{40} & \\ \hline &5&20&\color{orangered}{39}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 39 } = \color{blue}{ 78 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&5&10&-1&-18\\& & 10& 40& \color{blue}{78} \\ \hline &5&20&\color{blue}{39}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ 78 } = \color{orangered}{ 60 } $
$$ \begin{array}{c|rrrr}2&5&10&-1&\color{orangered}{ -18 }\\& & 10& 40& \color{orangered}{78} \\ \hline &\color{blue}{5}&\color{blue}{20}&\color{blue}{39}&\color{orangered}{60} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{2}+20x+39 } $ with a remainder of $ \color{red}{ 60 } $.