Divide using synthetic division $$ ( -x^{4}+2x^{3}+12x^{2}-16 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&-1&2&12&0&-16\\& & 2& -8& -8& \color{black}{16} \\ \hline &\color{blue}{-1}&\color{blue}{4}&\color{blue}{4}&\color{blue}{-8}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ -x^{4}+2x^{3}+12x^{2}-16 }{ x+2 } = \color{blue}{-x^{3}+4x^{2}+4x-8} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&-1&2&12&0&-16\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ -1 }&2&12&0&-16\\& & & & & \\ \hline &\color{orangered}{-1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&-1&2&12&0&-16\\& & \color{blue}{2} & & & \\ \hline &\color{blue}{-1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 2 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}-2&-1&\color{orangered}{ 2 }&12&0&-16\\& & \color{orangered}{2} & & & \\ \hline &-1&\color{orangered}{4}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 4 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&-1&2&12&0&-16\\& & 2& \color{blue}{-8} & & \\ \hline &-1&\color{blue}{4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}-2&-1&2&\color{orangered}{ 12 }&0&-16\\& & 2& \color{orangered}{-8} & & \\ \hline &-1&4&\color{orangered}{4}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 4 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&-1&2&12&0&-16\\& & 2& -8& \color{blue}{-8} & \\ \hline &-1&4&\color{blue}{4}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}-2&-1&2&12&\color{orangered}{ 0 }&-16\\& & 2& -8& \color{orangered}{-8} & \\ \hline &-1&4&4&\color{orangered}{-8}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&-1&2&12&0&-16\\& & 2& -8& -8& \color{blue}{16} \\ \hline &-1&4&4&\color{blue}{-8}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 16 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-2&-1&2&12&0&\color{orangered}{ -16 }\\& & 2& -8& -8& \color{orangered}{16} \\ \hline &\color{blue}{-1}&\color{blue}{4}&\color{blue}{4}&\color{blue}{-8}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -x^{3}+4x^{2}+4x-8 } $ with a remainder of $ \color{red}{ 0 } $.