Divide using synthetic division $$ ( x^{4}-7x^{3}-4x+12 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&1&-7&0&-4&12\\& & 3& -12& -36& \color{black}{-120} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{-12}&\color{blue}{-40}&\color{orangered}{-108} \end{array} $$The solution is:
$$ \frac{ x^{4}-7x^{3}-4x+12 }{ x-3 } = \color{blue}{x^{3}-4x^{2}-12x-40} \color{red}{~-~} \frac{ \color{red}{ 108 } }{ x-3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-7&0&-4&12\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 1 }&-7&0&-4&12\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-7&0&-4&12\\& & \color{blue}{3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 3 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}3&1&\color{orangered}{ -7 }&0&-4&12\\& & \color{orangered}{3} & & & \\ \hline &1&\color{orangered}{-4}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-7&0&-4&12\\& & 3& \color{blue}{-12} & & \\ \hline &1&\color{blue}{-4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrrr}3&1&-7&\color{orangered}{ 0 }&-4&12\\& & 3& \color{orangered}{-12} & & \\ \hline &1&-4&\color{orangered}{-12}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ -36 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-7&0&-4&12\\& & 3& -12& \color{blue}{-36} & \\ \hline &1&-4&\color{blue}{-12}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ \left( -36 \right) } = \color{orangered}{ -40 } $
$$ \begin{array}{c|rrrrr}3&1&-7&0&\color{orangered}{ -4 }&12\\& & 3& -12& \color{orangered}{-36} & \\ \hline &1&-4&-12&\color{orangered}{-40}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -40 \right) } = \color{blue}{ -120 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-7&0&-4&12\\& & 3& -12& -36& \color{blue}{-120} \\ \hline &1&-4&-12&\color{blue}{-40}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -120 \right) } = \color{orangered}{ -108 } $
$$ \begin{array}{c|rrrrr}3&1&-7&0&-4&\color{orangered}{ 12 }\\& & 3& -12& -36& \color{orangered}{-120} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{-12}&\color{blue}{-40}&\color{orangered}{-108} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-4x^{2}-12x-40 } $ with a remainder of $ \color{red}{ -108 } $.