Divide using synthetic division $$ ( 2x^{5}+7x^{3}-3x^{2}+12 ) \div ( x+1 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-1&2&0&7&-3&0&12\\& & -2& 2& -9& 12& \color{black}{-12} \\ \hline &\color{blue}{2}&\color{blue}{-2}&\color{blue}{9}&\color{blue}{-12}&\color{blue}{12}&\color{orangered}{0} \end{array} $$The solution is:
$$ \dfrac{ 2x^{5}+7x^{3}-3x^{2}+12 }{ x+1 } = \color{blue}{2x^{4}-2x^{3}+9x^{2}-12x+12} $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&2&0&7&-3&0&12\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-1&\color{orangered}{ 2 }&0&7&-3&0&12\\& & & & & & \\ \hline &\color{orangered}{2}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 2 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&2&0&7&-3&0&12\\& & \color{blue}{-2} & & & & \\ \hline &\color{blue}{2}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrrr}-1&2&\color{orangered}{ 0 }&7&-3&0&12\\& & \color{orangered}{-2} & & & & \\ \hline &2&\color{orangered}{-2}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&2&0&7&-3&0&12\\& & -2& \color{blue}{2} & & & \\ \hline &2&\color{blue}{-2}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 2 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrrr}-1&2&0&\color{orangered}{ 7 }&-3&0&12\\& & -2& \color{orangered}{2} & & & \\ \hline &2&-2&\color{orangered}{9}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 9 } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&2&0&7&-3&0&12\\& & -2& 2& \color{blue}{-9} & & \\ \hline &2&-2&\color{blue}{9}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrrrr}-1&2&0&7&\color{orangered}{ -3 }&0&12\\& & -2& 2& \color{orangered}{-9} & & \\ \hline &2&-2&9&\color{orangered}{-12}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&2&0&7&-3&0&12\\& & -2& 2& -9& \color{blue}{12} & \\ \hline &2&-2&9&\color{blue}{-12}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 12 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrrr}-1&2&0&7&-3&\color{orangered}{ 0 }&12\\& & -2& 2& -9& \color{orangered}{12} & \\ \hline &2&-2&9&-12&\color{orangered}{12}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 12 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&2&0&7&-3&0&12\\& & -2& 2& -9& 12& \color{blue}{-12} \\ \hline &2&-2&9&-12&\color{blue}{12}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}-1&2&0&7&-3&0&\color{orangered}{ 12 }\\& & -2& 2& -9& 12& \color{orangered}{-12} \\ \hline &\color{blue}{2}&\color{blue}{-2}&\color{blue}{9}&\color{blue}{-12}&\color{blue}{12}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{4}-2x^{3}+9x^{2}-12x+12 } $ with a remainder of $ \color{red}{ 0 } $.