Divide using synthetic division $$ ( 4x^{5}-6x^{4}+3x^{2}-9x+12 ) \div ( x+1 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-1&4&-6&0&3&-9&12\\& & -4& 10& -10& 7& \color{black}{2} \\ \hline &\color{blue}{4}&\color{blue}{-10}&\color{blue}{10}&\color{blue}{-7}&\color{blue}{-2}&\color{orangered}{14} \end{array} $$The solution is:
$$ \dfrac{ 4x^{5}-6x^{4}+3x^{2}-9x+12 }{ x+1 } = \color{blue}{4x^{4}-10x^{3}+10x^{2}-7x-2} ~+~ \dfrac{ \color{red}{ 14 } }{ x+1 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&4&-6&0&3&-9&12\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-1&\color{orangered}{ 4 }&-6&0&3&-9&12\\& & & & & & \\ \hline &\color{orangered}{4}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 4 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&4&-6&0&3&-9&12\\& & \color{blue}{-4} & & & & \\ \hline &\color{blue}{4}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrrr}-1&4&\color{orangered}{ -6 }&0&3&-9&12\\& & \color{orangered}{-4} & & & & \\ \hline &4&\color{orangered}{-10}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&4&-6&0&3&-9&12\\& & -4& \color{blue}{10} & & & \\ \hline &4&\color{blue}{-10}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 10 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrrr}-1&4&-6&\color{orangered}{ 0 }&3&-9&12\\& & -4& \color{orangered}{10} & & & \\ \hline &4&-10&\color{orangered}{10}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 10 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&4&-6&0&3&-9&12\\& & -4& 10& \color{blue}{-10} & & \\ \hline &4&-10&\color{blue}{10}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrrr}-1&4&-6&0&\color{orangered}{ 3 }&-9&12\\& & -4& 10& \color{orangered}{-10} & & \\ \hline &4&-10&10&\color{orangered}{-7}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 7 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&4&-6&0&3&-9&12\\& & -4& 10& -10& \color{blue}{7} & \\ \hline &4&-10&10&\color{blue}{-7}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ 7 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrrr}-1&4&-6&0&3&\color{orangered}{ -9 }&12\\& & -4& 10& -10& \color{orangered}{7} & \\ \hline &4&-10&10&-7&\color{orangered}{-2}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&4&-6&0&3&-9&12\\& & -4& 10& -10& 7& \color{blue}{2} \\ \hline &4&-10&10&-7&\color{blue}{-2}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 2 } = \color{orangered}{ 14 } $
$$ \begin{array}{c|rrrrrr}-1&4&-6&0&3&-9&\color{orangered}{ 12 }\\& & -4& 10& -10& 7& \color{orangered}{2} \\ \hline &\color{blue}{4}&\color{blue}{-10}&\color{blue}{10}&\color{blue}{-7}&\color{blue}{-2}&\color{orangered}{14} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{4}-10x^{3}+10x^{2}-7x-2 } $ with a remainder of $ \color{red}{ 14 } $.