Divide using synthetic division $$ ( 4x^{3}+20x^{2}-x-5 ) \div ( 2x+1 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrr}-1&4&20&-1&-5\\& & -4& -16& \color{black}{17} \\ \hline &\color{blue}{4}&\color{blue}{16}&\color{blue}{-17}&\color{orangered}{12} \end{array} $$The solution is:
$$ \dfrac{ 4x^{3}+20x^{2}-x-5 }{ x+1 } = \color{blue}{4x^{2}+16x-17} ~+~ \dfrac{ \color{red}{ 12 } }{ x+1 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 1 = 0 $ ( $ x = \color{blue}{ -1 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&4&20&-1&-5\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-1&\color{orangered}{ 4 }&20&-1&-5\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 4 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&4&20&-1&-5\\& & \color{blue}{-4} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrrr}-1&4&\color{orangered}{ 20 }&-1&-5\\& & \color{orangered}{-4} & & \\ \hline &4&\color{orangered}{16}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 16 } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&4&20&-1&-5\\& & -4& \color{blue}{-16} & \\ \hline &4&\color{blue}{16}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -17 } $
$$ \begin{array}{c|rrrr}-1&4&20&\color{orangered}{ -1 }&-5\\& & -4& \color{orangered}{-16} & \\ \hline &4&16&\color{orangered}{-17}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -17 \right) } = \color{blue}{ 17 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&4&20&-1&-5\\& & -4& -16& \color{blue}{17} \\ \hline &4&16&\color{blue}{-17}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 17 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrr}-1&4&20&-1&\color{orangered}{ -5 }\\& & -4& -16& \color{orangered}{17} \\ \hline &\color{blue}{4}&\color{blue}{16}&\color{blue}{-17}&\color{orangered}{12} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}+16x-17 } $ with a remainder of $ \color{red}{ 12 } $.