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