Divide using synthetic division $$ ( 6x^{6}+2x^{5}+3x^{4}+10 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrrr}3&6&2&3&0&0&0&10\\& & 18& 60& 189& 567& 1701& \color{black}{5103} \\ \hline &\color{blue}{6}&\color{blue}{20}&\color{blue}{63}&\color{blue}{189}&\color{blue}{567}&\color{blue}{1701}&\color{orangered}{5113} \end{array} $$The solution is:
$$ \frac{ 6x^{6}+2x^{5}+3x^{4}+10 }{ x-3 } = \color{blue}{6x^{5}+20x^{4}+63x^{3}+189x^{2}+567x+1701} ~+~ \frac{ \color{red}{ 5113 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrrrr}\color{blue}{3}&6&2&3&0&0&0&10\\& & & & & & & \\ \hline &&&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrrr}3&\color{orangered}{ 6 }&2&3&0&0&0&10\\& & & & & & & \\ \hline &\color{orangered}{6}&&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 6 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{3}&6&2&3&0&0&0&10\\& & \color{blue}{18} & & & & & \\ \hline &\color{blue}{6}&&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 18 } = \color{orangered}{ 20 } $
$$ \begin{array}{c|rrrrrrr}3&6&\color{orangered}{ 2 }&3&0&0&0&10\\& & \color{orangered}{18} & & & & & \\ \hline &6&\color{orangered}{20}&&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 20 } = \color{blue}{ 60 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{3}&6&2&3&0&0&0&10\\& & 18& \color{blue}{60} & & & & \\ \hline &6&\color{blue}{20}&&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 60 } = \color{orangered}{ 63 } $
$$ \begin{array}{c|rrrrrrr}3&6&2&\color{orangered}{ 3 }&0&0&0&10\\& & 18& \color{orangered}{60} & & & & \\ \hline &6&20&\color{orangered}{63}&&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 63 } = \color{blue}{ 189 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{3}&6&2&3&0&0&0&10\\& & 18& 60& \color{blue}{189} & & & \\ \hline &6&20&\color{blue}{63}&&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 189 } = \color{orangered}{ 189 } $
$$ \begin{array}{c|rrrrrrr}3&6&2&3&\color{orangered}{ 0 }&0&0&10\\& & 18& 60& \color{orangered}{189} & & & \\ \hline &6&20&63&\color{orangered}{189}&&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 189 } = \color{blue}{ 567 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{3}&6&2&3&0&0&0&10\\& & 18& 60& 189& \color{blue}{567} & & \\ \hline &6&20&63&\color{blue}{189}&&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 567 } = \color{orangered}{ 567 } $
$$ \begin{array}{c|rrrrrrr}3&6&2&3&0&\color{orangered}{ 0 }&0&10\\& & 18& 60& 189& \color{orangered}{567} & & \\ \hline &6&20&63&189&\color{orangered}{567}&& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 567 } = \color{blue}{ 1701 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{3}&6&2&3&0&0&0&10\\& & 18& 60& 189& 567& \color{blue}{1701} & \\ \hline &6&20&63&189&\color{blue}{567}&& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 1701 } = \color{orangered}{ 1701 } $
$$ \begin{array}{c|rrrrrrr}3&6&2&3&0&0&\color{orangered}{ 0 }&10\\& & 18& 60& 189& 567& \color{orangered}{1701} & \\ \hline &6&20&63&189&567&\color{orangered}{1701}& \end{array} $$Step 12 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1701 } = \color{blue}{ 5103 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{3}&6&2&3&0&0&0&10\\& & 18& 60& 189& 567& 1701& \color{blue}{5103} \\ \hline &6&20&63&189&567&\color{blue}{1701}& \end{array} $$Step 13 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ 5103 } = \color{orangered}{ 5113 } $
$$ \begin{array}{c|rrrrrrr}3&6&2&3&0&0&0&\color{orangered}{ 10 }\\& & 18& 60& 189& 567& 1701& \color{orangered}{5103} \\ \hline &\color{blue}{6}&\color{blue}{20}&\color{blue}{63}&\color{blue}{189}&\color{blue}{567}&\color{blue}{1701}&\color{orangered}{5113} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{5}+20x^{4}+63x^{3}+189x^{2}+567x+1701 } $ with a remainder of $ \color{red}{ 5113 } $.