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