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