Divide using synthetic division $$ ( 7x^{3}+3x^{2}+x-1 ) \div ( x+2 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&7&3&1&-1\\& & -14& 22& \color{black}{-46} \\ \hline &\color{blue}{7}&\color{blue}{-11}&\color{blue}{23}&\color{orangered}{-47} \end{array} $$The solution is:
$$ \dfrac{ 7x^{3}+3x^{2}+x-1 }{ x+2 } = \color{blue}{7x^{2}-11x+23} \color{red}{~-~} \dfrac{ \color{red}{ 47 } }{ x+2 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&7&3&1&-1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 7 }&3&1&-1\\& & & & \\ \hline &\color{orangered}{7}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 7 } = \color{blue}{ -14 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&7&3&1&-1\\& & \color{blue}{-14} & & \\ \hline &\color{blue}{7}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -14 \right) } = \color{orangered}{ -11 } $
$$ \begin{array}{c|rrrr}-2&7&\color{orangered}{ 3 }&1&-1\\& & \color{orangered}{-14} & & \\ \hline &7&\color{orangered}{-11}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -11 \right) } = \color{blue}{ 22 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&7&3&1&-1\\& & -14& \color{blue}{22} & \\ \hline &7&\color{blue}{-11}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 22 } = \color{orangered}{ 23 } $
$$ \begin{array}{c|rrrr}-2&7&3&\color{orangered}{ 1 }&-1\\& & -14& \color{orangered}{22} & \\ \hline &7&-11&\color{orangered}{23}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 23 } = \color{blue}{ -46 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&7&3&1&-1\\& & -14& 22& \color{blue}{-46} \\ \hline &7&-11&\color{blue}{23}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -46 \right) } = \color{orangered}{ -47 } $
$$ \begin{array}{c|rrrr}-2&7&3&1&\color{orangered}{ -1 }\\& & -14& 22& \color{orangered}{-46} \\ \hline &\color{blue}{7}&\color{blue}{-11}&\color{blue}{23}&\color{orangered}{-47} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 7x^{2}-11x+23 } $ with a remainder of $ \color{red}{ -47 } $.