Divide using synthetic division $$ ( x^{5}+16x^{4}+21x^{3}+6x^{2}+10x+6 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}3&1&16&21&6&10&6\\& & 3& 57& 234& 720& \color{black}{2190} \\ \hline &\color{blue}{1}&\color{blue}{19}&\color{blue}{78}&\color{blue}{240}&\color{blue}{730}&\color{orangered}{2196} \end{array} $$The solution is:
$$ \frac{ x^{5}+16x^{4}+21x^{3}+6x^{2}+10x+6 }{ x-3 } = \color{blue}{x^{4}+19x^{3}+78x^{2}+240x+730} ~+~ \frac{ \color{red}{ 2196 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&16&21&6&10&6\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}3&\color{orangered}{ 1 }&16&21&6&10&6\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&16&21&6&10&6\\& & \color{blue}{3} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ 3 } = \color{orangered}{ 19 } $
$$ \begin{array}{c|rrrrrr}3&1&\color{orangered}{ 16 }&21&6&10&6\\& & \color{orangered}{3} & & & & \\ \hline &1&\color{orangered}{19}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 19 } = \color{blue}{ 57 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&16&21&6&10&6\\& & 3& \color{blue}{57} & & & \\ \hline &1&\color{blue}{19}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 21 } + \color{orangered}{ 57 } = \color{orangered}{ 78 } $
$$ \begin{array}{c|rrrrrr}3&1&16&\color{orangered}{ 21 }&6&10&6\\& & 3& \color{orangered}{57} & & & \\ \hline &1&19&\color{orangered}{78}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 78 } = \color{blue}{ 234 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&16&21&6&10&6\\& & 3& 57& \color{blue}{234} & & \\ \hline &1&19&\color{blue}{78}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 234 } = \color{orangered}{ 240 } $
$$ \begin{array}{c|rrrrrr}3&1&16&21&\color{orangered}{ 6 }&10&6\\& & 3& 57& \color{orangered}{234} & & \\ \hline &1&19&78&\color{orangered}{240}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 240 } = \color{blue}{ 720 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&16&21&6&10&6\\& & 3& 57& 234& \color{blue}{720} & \\ \hline &1&19&78&\color{blue}{240}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ 720 } = \color{orangered}{ 730 } $
$$ \begin{array}{c|rrrrrr}3&1&16&21&6&\color{orangered}{ 10 }&6\\& & 3& 57& 234& \color{orangered}{720} & \\ \hline &1&19&78&240&\color{orangered}{730}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 730 } = \color{blue}{ 2190 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&16&21&6&10&6\\& & 3& 57& 234& 720& \color{blue}{2190} \\ \hline &1&19&78&240&\color{blue}{730}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 2190 } = \color{orangered}{ 2196 } $
$$ \begin{array}{c|rrrrrr}3&1&16&21&6&10&\color{orangered}{ 6 }\\& & 3& 57& 234& 720& \color{orangered}{2190} \\ \hline &\color{blue}{1}&\color{blue}{19}&\color{blue}{78}&\color{blue}{240}&\color{blue}{730}&\color{orangered}{2196} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}+19x^{3}+78x^{2}+240x+730 } $ with a remainder of $ \color{red}{ 2196 } $.