Divide using synthetic division $$ ( x^{4}-5x^{3}+5x^{2}+7x-12 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}4&1&-5&5&7&-12\\& & 4& -4& 4& \color{black}{44} \\ \hline &\color{blue}{1}&\color{blue}{-1}&\color{blue}{1}&\color{blue}{11}&\color{orangered}{32} \end{array} $$The solution is:
$$ \frac{ x^{4}-5x^{3}+5x^{2}+7x-12 }{ x-4 } = \color{blue}{x^{3}-x^{2}+x+11} ~+~ \frac{ \color{red}{ 32 } }{ x-4 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&-5&5&7&-12\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}4&\color{orangered}{ 1 }&-5&5&7&-12\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 1 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&-5&5&7&-12\\& & \color{blue}{4} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 4 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}4&1&\color{orangered}{ -5 }&5&7&-12\\& & \color{orangered}{4} & & & \\ \hline &1&\color{orangered}{-1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&-5&5&7&-12\\& & 4& \color{blue}{-4} & & \\ \hline &1&\color{blue}{-1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}4&1&-5&\color{orangered}{ 5 }&7&-12\\& & 4& \color{orangered}{-4} & & \\ \hline &1&-1&\color{orangered}{1}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 1 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&-5&5&7&-12\\& & 4& -4& \color{blue}{4} & \\ \hline &1&-1&\color{blue}{1}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 4 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrrr}4&1&-5&5&\color{orangered}{ 7 }&-12\\& & 4& -4& \color{orangered}{4} & \\ \hline &1&-1&1&\color{orangered}{11}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 11 } = \color{blue}{ 44 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&-5&5&7&-12\\& & 4& -4& 4& \color{blue}{44} \\ \hline &1&-1&1&\color{blue}{11}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 44 } = \color{orangered}{ 32 } $
$$ \begin{array}{c|rrrrr}4&1&-5&5&7&\color{orangered}{ -12 }\\& & 4& -4& 4& \color{orangered}{44} \\ \hline &\color{blue}{1}&\color{blue}{-1}&\color{blue}{1}&\color{blue}{11}&\color{orangered}{32} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-x^{2}+x+11 } $ with a remainder of $ \color{red}{ 32 } $.