Divide using synthetic division $$ ( 3x^{4}+8x^{3}+21x^{2}+26x+6 ) \div ( 3x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&3&8&21&26&6\\& & 3& 11& 32& \color{black}{58} \\ \hline &\color{blue}{3}&\color{blue}{11}&\color{blue}{32}&\color{blue}{58}&\color{orangered}{64} \end{array} $$The solution is:
$$ \frac{ 3x^{4}+8x^{3}+21x^{2}+26x+6 }{ x-1 } = \color{blue}{3x^{3}+11x^{2}+32x+58} ~+~ \frac{ \color{red}{ 64 } }{ x-1 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&3&8&21&26&6\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 3 }&8&21&26&6\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 3 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&3&8&21&26&6\\& & \color{blue}{3} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 3 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrrr}1&3&\color{orangered}{ 8 }&21&26&6\\& & \color{orangered}{3} & & & \\ \hline &3&\color{orangered}{11}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 11 } = \color{blue}{ 11 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&3&8&21&26&6\\& & 3& \color{blue}{11} & & \\ \hline &3&\color{blue}{11}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 21 } + \color{orangered}{ 11 } = \color{orangered}{ 32 } $
$$ \begin{array}{c|rrrrr}1&3&8&\color{orangered}{ 21 }&26&6\\& & 3& \color{orangered}{11} & & \\ \hline &3&11&\color{orangered}{32}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 32 } = \color{blue}{ 32 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&3&8&21&26&6\\& & 3& 11& \color{blue}{32} & \\ \hline &3&11&\color{blue}{32}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 26 } + \color{orangered}{ 32 } = \color{orangered}{ 58 } $
$$ \begin{array}{c|rrrrr}1&3&8&21&\color{orangered}{ 26 }&6\\& & 3& 11& \color{orangered}{32} & \\ \hline &3&11&32&\color{orangered}{58}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 58 } = \color{blue}{ 58 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&3&8&21&26&6\\& & 3& 11& 32& \color{blue}{58} \\ \hline &3&11&32&\color{blue}{58}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 58 } = \color{orangered}{ 64 } $
$$ \begin{array}{c|rrrrr}1&3&8&21&26&\color{orangered}{ 6 }\\& & 3& 11& 32& \color{orangered}{58} \\ \hline &\color{blue}{3}&\color{blue}{11}&\color{blue}{32}&\color{blue}{58}&\color{orangered}{64} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}+11x^{2}+32x+58 } $ with a remainder of $ \color{red}{ 64 } $.