Divide using synthetic division $$ ( 6x^{3}-7x^{2}-15x+1 ) \div ( 3x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-1&6&-7&-15&1\\& & -6& 13& \color{black}{2} \\ \hline &\color{blue}{6}&\color{blue}{-13}&\color{blue}{-2}&\color{orangered}{3} \end{array} $$The solution is:
$$ \frac{ 6x^{3}-7x^{2}-15x+1 }{ x+1 } = \color{blue}{6x^{2}-13x-2} ~+~ \frac{ \color{red}{ 3 } }{ x+1 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&6&-7&-15&1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-1&\color{orangered}{ 6 }&-7&-15&1\\& & & & \\ \hline &\color{orangered}{6}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&6&-7&-15&1\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -13 } $
$$ \begin{array}{c|rrrr}-1&6&\color{orangered}{ -7 }&-15&1\\& & \color{orangered}{-6} & & \\ \hline &6&\color{orangered}{-13}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -13 \right) } = \color{blue}{ 13 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&6&-7&-15&1\\& & -6& \color{blue}{13} & \\ \hline &6&\color{blue}{-13}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 13 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}-1&6&-7&\color{orangered}{ -15 }&1\\& & -6& \color{orangered}{13} & \\ \hline &6&-13&\color{orangered}{-2}& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&6&-7&-15&1\\& & -6& 13& \color{blue}{2} \\ \hline &6&-13&\color{blue}{-2}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 2 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}-1&6&-7&-15&\color{orangered}{ 1 }\\& & -6& 13& \color{orangered}{2} \\ \hline &\color{blue}{6}&\color{blue}{-13}&\color{blue}{-2}&\color{orangered}{3} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{2}-13x-2 } $ with a remainder of $ \color{red}{ 3 } $.