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