Divide using synthetic division $$ ( 8x^{3}+4x^{2}-4x-8 ) \div ( 4x+2 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&8&4&-4&-8\\& & -16& 24& \color{black}{-40} \\ \hline &\color{blue}{8}&\color{blue}{-12}&\color{blue}{20}&\color{orangered}{-48} \end{array} $$The solution is:
$$ \dfrac{ 8x^{3}+4x^{2}-4x-8 }{ x+2 } = \color{blue}{8x^{2}-12x+20} \color{red}{~-~} \dfrac{ \color{red}{ 48 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&8&4&-4&-8\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 8 }&4&-4&-8\\& & & & \\ \hline &\color{orangered}{8}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 8 } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&8&4&-4&-8\\& & \color{blue}{-16} & & \\ \hline &\color{blue}{8}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrr}-2&8&\color{orangered}{ 4 }&-4&-8\\& & \color{orangered}{-16} & & \\ \hline &8&\color{orangered}{-12}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&8&4&-4&-8\\& & -16& \color{blue}{24} & \\ \hline &8&\color{blue}{-12}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 24 } = \color{orangered}{ 20 } $
$$ \begin{array}{c|rrrr}-2&8&4&\color{orangered}{ -4 }&-8\\& & -16& \color{orangered}{24} & \\ \hline &8&-12&\color{orangered}{20}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 20 } = \color{blue}{ -40 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&8&4&-4&-8\\& & -16& 24& \color{blue}{-40} \\ \hline &8&-12&\color{blue}{20}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ \left( -40 \right) } = \color{orangered}{ -48 } $
$$ \begin{array}{c|rrrr}-2&8&4&-4&\color{orangered}{ -8 }\\& & -16& 24& \color{orangered}{-40} \\ \hline &\color{blue}{8}&\color{blue}{-12}&\color{blue}{20}&\color{orangered}{-48} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 8x^{2}-12x+20 } $ with a remainder of $ \color{red}{ -48 } $.