Divide using synthetic division $$ ( x^{4}-6561 ) \div ( x-9 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}9&1&0&0&0&-6561\\& & 9& 81& 729& \color{black}{6561} \\ \hline &\color{blue}{1}&\color{blue}{9}&\color{blue}{81}&\color{blue}{729}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{4}-6561 }{ x-9 } = \color{blue}{x^{3}+9x^{2}+81x+729} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -9 = 0 $ ( $ x = \color{blue}{ 9 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{9}&1&0&0&0&-6561\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}9&\color{orangered}{ 1 }&0&0&0&-6561\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 9 } \cdot \color{blue}{ 1 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{9}&1&0&0&0&-6561\\& & \color{blue}{9} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 9 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrr}9&1&\color{orangered}{ 0 }&0&0&-6561\\& & \color{orangered}{9} & & & \\ \hline &1&\color{orangered}{9}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 9 } \cdot \color{blue}{ 9 } = \color{blue}{ 81 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{9}&1&0&0&0&-6561\\& & 9& \color{blue}{81} & & \\ \hline &1&\color{blue}{9}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 81 } = \color{orangered}{ 81 } $
$$ \begin{array}{c|rrrrr}9&1&0&\color{orangered}{ 0 }&0&-6561\\& & 9& \color{orangered}{81} & & \\ \hline &1&9&\color{orangered}{81}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 9 } \cdot \color{blue}{ 81 } = \color{blue}{ 729 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{9}&1&0&0&0&-6561\\& & 9& 81& \color{blue}{729} & \\ \hline &1&9&\color{blue}{81}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 729 } = \color{orangered}{ 729 } $
$$ \begin{array}{c|rrrrr}9&1&0&0&\color{orangered}{ 0 }&-6561\\& & 9& 81& \color{orangered}{729} & \\ \hline &1&9&81&\color{orangered}{729}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 9 } \cdot \color{blue}{ 729 } = \color{blue}{ 6561 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{9}&1&0&0&0&-6561\\& & 9& 81& 729& \color{blue}{6561} \\ \hline &1&9&81&\color{blue}{729}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -6561 } + \color{orangered}{ 6561 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}9&1&0&0&0&\color{orangered}{ -6561 }\\& & 9& 81& 729& \color{orangered}{6561} \\ \hline &\color{blue}{1}&\color{blue}{9}&\color{blue}{81}&\color{blue}{729}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+9x^{2}+81x+729 } $ with a remainder of $ \color{red}{ 0 } $.