Divide using synthetic division $$ ( x^{5}+6x^{4}+5x^{3}+16x^{2}-12x-16 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}1&1&6&5&16&-12&-16\\& & 1& 7& 12& 28& \color{black}{16} \\ \hline &\color{blue}{1}&\color{blue}{7}&\color{blue}{12}&\color{blue}{28}&\color{blue}{16}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{5}+6x^{4}+5x^{3}+16x^{2}-12x-16 }{ x-1 } = \color{blue}{x^{4}+7x^{3}+12x^{2}+28x+16} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&1&6&5&16&-12&-16\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}1&\color{orangered}{ 1 }&6&5&16&-12&-16\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 1 } = \color{blue}{ 1 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&1&6&5&16&-12&-16\\& & \color{blue}{1} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 1 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrrr}1&1&\color{orangered}{ 6 }&5&16&-12&-16\\& & \color{orangered}{1} & & & & \\ \hline &1&\color{orangered}{7}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 7 } = \color{blue}{ 7 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&1&6&5&16&-12&-16\\& & 1& \color{blue}{7} & & & \\ \hline &1&\color{blue}{7}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 7 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrrr}1&1&6&\color{orangered}{ 5 }&16&-12&-16\\& & 1& \color{orangered}{7} & & & \\ \hline &1&7&\color{orangered}{12}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 12 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&1&6&5&16&-12&-16\\& & 1& 7& \color{blue}{12} & & \\ \hline &1&7&\color{blue}{12}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ 12 } = \color{orangered}{ 28 } $
$$ \begin{array}{c|rrrrrr}1&1&6&5&\color{orangered}{ 16 }&-12&-16\\& & 1& 7& \color{orangered}{12} & & \\ \hline &1&7&12&\color{orangered}{28}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 28 } = \color{blue}{ 28 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&1&6&5&16&-12&-16\\& & 1& 7& 12& \color{blue}{28} & \\ \hline &1&7&12&\color{blue}{28}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 28 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrrrrr}1&1&6&5&16&\color{orangered}{ -12 }&-16\\& & 1& 7& 12& \color{orangered}{28} & \\ \hline &1&7&12&28&\color{orangered}{16}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 16 } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&1&6&5&16&-12&-16\\& & 1& 7& 12& 28& \color{blue}{16} \\ \hline &1&7&12&28&\color{blue}{16}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 16 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}1&1&6&5&16&-12&\color{orangered}{ -16 }\\& & 1& 7& 12& 28& \color{orangered}{16} \\ \hline &\color{blue}{1}&\color{blue}{7}&\color{blue}{12}&\color{blue}{28}&\color{blue}{16}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}+7x^{3}+12x^{2}+28x+16 } $ with a remainder of $ \color{red}{ 0 } $.