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