Divide using synthetic division $$ ( x^{5}-2x^{4}-x^{3}-3 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}3&1&-2&-1&0&0&-3\\& & 3& 3& 6& 18& \color{black}{54} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{2}&\color{blue}{6}&\color{blue}{18}&\color{orangered}{51} \end{array} $$The solution is:
$$ \frac{ x^{5}-2x^{4}-x^{3}-3 }{ x-3 } = \color{blue}{x^{4}+x^{3}+2x^{2}+6x+18} ~+~ \frac{ \color{red}{ 51 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-2&-1&0&0&-3\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}3&\color{orangered}{ 1 }&-2&-1&0&0&-3\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-2&-1&0&0&-3\\& & \color{blue}{3} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 3 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrrr}3&1&\color{orangered}{ -2 }&-1&0&0&-3\\& & \color{orangered}{3} & & & & \\ \hline &1&\color{orangered}{1}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-2&-1&0&0&-3\\& & 3& \color{blue}{3} & & & \\ \hline &1&\color{blue}{1}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 3 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrrr}3&1&-2&\color{orangered}{ -1 }&0&0&-3\\& & 3& \color{orangered}{3} & & & \\ \hline &1&1&\color{orangered}{2}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 2 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-2&-1&0&0&-3\\& & 3& 3& \color{blue}{6} & & \\ \hline &1&1&\color{blue}{2}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 6 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrrr}3&1&-2&-1&\color{orangered}{ 0 }&0&-3\\& & 3& 3& \color{orangered}{6} & & \\ \hline &1&1&2&\color{orangered}{6}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 6 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-2&-1&0&0&-3\\& & 3& 3& 6& \color{blue}{18} & \\ \hline &1&1&2&\color{blue}{6}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 18 } = \color{orangered}{ 18 } $
$$ \begin{array}{c|rrrrrr}3&1&-2&-1&0&\color{orangered}{ 0 }&-3\\& & 3& 3& 6& \color{orangered}{18} & \\ \hline &1&1&2&6&\color{orangered}{18}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 18 } = \color{blue}{ 54 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-2&-1&0&0&-3\\& & 3& 3& 6& 18& \color{blue}{54} \\ \hline &1&1&2&6&\color{blue}{18}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 54 } = \color{orangered}{ 51 } $
$$ \begin{array}{c|rrrrrr}3&1&-2&-1&0&0&\color{orangered}{ -3 }\\& & 3& 3& 6& 18& \color{orangered}{54} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{2}&\color{blue}{6}&\color{blue}{18}&\color{orangered}{51} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}+x^{3}+2x^{2}+6x+18 } $ with a remainder of $ \color{red}{ 51 } $.