Divide using synthetic division $$ ( -x^{3}-6x^{2}+2x+4 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&-1&-6&2&4\\& & 2& 8& \color{black}{-20} \\ \hline &\color{blue}{-1}&\color{blue}{-4}&\color{blue}{10}&\color{orangered}{-16} \end{array} $$The solution is:
$$ \frac{ -x^{3}-6x^{2}+2x+4 }{ x+2 } = \color{blue}{-x^{2}-4x+10} \color{red}{~-~} \frac{ \color{red}{ 16 } }{ x+2 } $$Explanation
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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&-1&-6&2&4\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ -1 }&-6&2&4\\& & & & \\ \hline &\color{orangered}{-1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&-1&-6&2&4\\& & \color{blue}{2} & & \\ \hline &\color{blue}{-1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 2 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}-2&-1&\color{orangered}{ -6 }&2&4\\& & \color{orangered}{2} & & \\ \hline &-1&\color{orangered}{-4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&-1&-6&2&4\\& & 2& \color{blue}{8} & \\ \hline &-1&\color{blue}{-4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 8 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrr}-2&-1&-6&\color{orangered}{ 2 }&4\\& & 2& \color{orangered}{8} & \\ \hline &-1&-4&\color{orangered}{10}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 10 } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&-1&-6&2&4\\& & 2& 8& \color{blue}{-20} \\ \hline &-1&-4&\color{blue}{10}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrr}-2&-1&-6&2&\color{orangered}{ 4 }\\& & 2& 8& \color{orangered}{-20} \\ \hline &\color{blue}{-1}&\color{blue}{-4}&\color{blue}{10}&\color{orangered}{-16} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -x^{2}-4x+10 } $ with a remainder of $ \color{red}{ -16 } $.