Divide using synthetic division $$ ( x^{3}-5x^{2}+6 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&1&-5&0&6\\& & 4& -4& \color{black}{-16} \\ \hline &\color{blue}{1}&\color{blue}{-1}&\color{blue}{-4}&\color{orangered}{-10} \end{array} $$The solution is:
$$ \frac{ x^{3}-5x^{2}+6 }{ x-4 } = \color{blue}{x^{2}-x-4} \color{red}{~-~} \frac{ \color{red}{ 10 } }{ x-4 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&-5&0&6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 1 }&-5&0&6\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&-5&0&6\\& & \color{blue}{4} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 4 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}4&1&\color{orangered}{ -5 }&0&6\\& & \color{orangered}{4} & & \\ \hline &1&\color{orangered}{-1}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&-5&0&6\\& & 4& \color{blue}{-4} & \\ \hline &1&\color{blue}{-1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}4&1&-5&\color{orangered}{ 0 }&6\\& & 4& \color{orangered}{-4} & \\ \hline &1&-1&\color{orangered}{-4}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&-5&0&6\\& & 4& -4& \color{blue}{-16} \\ \hline &1&-1&\color{blue}{-4}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrr}4&1&-5&0&\color{orangered}{ 6 }\\& & 4& -4& \color{orangered}{-16} \\ \hline &\color{blue}{1}&\color{blue}{-1}&\color{blue}{-4}&\color{orangered}{-10} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-x-4 } $ with a remainder of $ \color{red}{ -10 } $.