Divide using synthetic division $$ ( x^{5}+x^{3}-x ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}3&1&0&1&0&-1&0\\& & 3& 9& 30& 90& \color{black}{267} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{10}&\color{blue}{30}&\color{blue}{89}&\color{orangered}{267} \end{array} $$The solution is:
$$ \frac{ x^{5}+x^{3}-x }{ x-3 } = \color{blue}{x^{4}+3x^{3}+10x^{2}+30x+89} ~+~ \frac{ \color{red}{ 267 } }{ x-3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&0&1&0&-1&0\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}3&\color{orangered}{ 1 }&0&1&0&-1&0\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&0&1&0&-1&0\\& & \color{blue}{3} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 3 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrrr}3&1&\color{orangered}{ 0 }&1&0&-1&0\\& & \color{orangered}{3} & & & & \\ \hline &1&\color{orangered}{3}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 3 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&0&1&0&-1&0\\& & 3& \color{blue}{9} & & & \\ \hline &1&\color{blue}{3}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 9 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrrr}3&1&0&\color{orangered}{ 1 }&0&-1&0\\& & 3& \color{orangered}{9} & & & \\ \hline &1&3&\color{orangered}{10}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 10 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&0&1&0&-1&0\\& & 3& 9& \color{blue}{30} & & \\ \hline &1&3&\color{blue}{10}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 30 } = \color{orangered}{ 30 } $
$$ \begin{array}{c|rrrrrr}3&1&0&1&\color{orangered}{ 0 }&-1&0\\& & 3& 9& \color{orangered}{30} & & \\ \hline &1&3&10&\color{orangered}{30}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 30 } = \color{blue}{ 90 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&0&1&0&-1&0\\& & 3& 9& 30& \color{blue}{90} & \\ \hline &1&3&10&\color{blue}{30}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 90 } = \color{orangered}{ 89 } $
$$ \begin{array}{c|rrrrrr}3&1&0&1&0&\color{orangered}{ -1 }&0\\& & 3& 9& 30& \color{orangered}{90} & \\ \hline &1&3&10&30&\color{orangered}{89}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 89 } = \color{blue}{ 267 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&0&1&0&-1&0\\& & 3& 9& 30& 90& \color{blue}{267} \\ \hline &1&3&10&30&\color{blue}{89}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 267 } = \color{orangered}{ 267 } $
$$ \begin{array}{c|rrrrrr}3&1&0&1&0&-1&\color{orangered}{ 0 }\\& & 3& 9& 30& 90& \color{orangered}{267} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{10}&\color{blue}{30}&\color{blue}{89}&\color{orangered}{267} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}+3x^{3}+10x^{2}+30x+89 } $ with a remainder of $ \color{red}{ 267 } $.