Divide using synthetic division $$ ( -x^{3}+6x^{2}-12x-16 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&-1&6&-12&-16\\& & -4& 8& \color{black}{-16} \\ \hline &\color{blue}{-1}&\color{blue}{2}&\color{blue}{-4}&\color{orangered}{-32} \end{array} $$The solution is:
$$ \frac{ -x^{3}+6x^{2}-12x-16 }{ x-4 } = \color{blue}{-x^{2}+2x-4} \color{red}{~-~} \frac{ \color{red}{ 32 } }{ x-4 } $$Explanation
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}{ \left( -4 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}4&-1&\color{orangered}{ 6 }&-12&-16\\& & \color{orangered}{-4} & & \\ \hline &-1&\color{orangered}{2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 2 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&-1&6&-12&-16\\& & -4& \color{blue}{8} & \\ \hline &-1&\color{blue}{2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 8 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}4&-1&6&\color{orangered}{ -12 }&-16\\& & -4& \color{orangered}{8} & \\ \hline &-1&2&\color{orangered}{-4}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&-1&6&-12&-16\\& & -4& 8& \color{blue}{-16} \\ \hline &-1&2&\color{blue}{-4}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -32 } $
$$ \begin{array}{c|rrrr}4&-1&6&-12&\color{orangered}{ -16 }\\& & -4& 8& \color{orangered}{-16} \\ \hline &\color{blue}{-1}&\color{blue}{2}&\color{blue}{-4}&\color{orangered}{-32} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -x^{2}+2x-4 } $ with a remainder of $ \color{red}{ -32 } $.