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