Divide using synthetic division $$ ( -x^{3}+6x^{2}-12x-16 ) \div ( x+4 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrr}-4&-1&6&-12&-16\\& & 4& -40& \color{black}{208} \\ \hline &\color{blue}{-1}&\color{blue}{10}&\color{blue}{-52}&\color{orangered}{192} \end{array} $$The solution is:
$$ \dfrac{ -x^{3}+6x^{2}-12x-16 }{ x+4 } = \color{blue}{-x^{2}+10x-52} ~+~ \dfrac{ \color{red}{ 192 } }{ x+4 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 4 = 0 $ ( $ x = \color{blue}{ -4 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&-1&6&-12&-16\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-4&\color{orangered}{ -1 }&6&-12&-16\\& & & & \\ \hline &\color{orangered}{-1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&-1&6&-12&-16\\& & \color{blue}{4} & & \\ \hline &\color{blue}{-1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 4 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrr}-4&-1&\color{orangered}{ 6 }&-12&-16\\& & \color{orangered}{4} & & \\ \hline &-1&\color{orangered}{10}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 10 } = \color{blue}{ -40 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&-1&6&-12&-16\\& & 4& \color{blue}{-40} & \\ \hline &-1&\color{blue}{10}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ \left( -40 \right) } = \color{orangered}{ -52 } $
$$ \begin{array}{c|rrrr}-4&-1&6&\color{orangered}{ -12 }&-16\\& & 4& \color{orangered}{-40} & \\ \hline &-1&10&\color{orangered}{-52}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -52 \right) } = \color{blue}{ 208 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&-1&6&-12&-16\\& & 4& -40& \color{blue}{208} \\ \hline &-1&10&\color{blue}{-52}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 208 } = \color{orangered}{ 192 } $
$$ \begin{array}{c|rrrr}-4&-1&6&-12&\color{orangered}{ -16 }\\& & 4& -40& \color{orangered}{208} \\ \hline &\color{blue}{-1}&\color{blue}{10}&\color{blue}{-52}&\color{orangered}{192} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -x^{2}+10x-52 } $ with a remainder of $ \color{red}{ 192 } $.