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