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