Divide using synthetic division $$ ( x^{3}-6x^{2}+4x+23 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&1&-6&4&23\\& & 4& -8& \color{black}{-16} \\ \hline &\color{blue}{1}&\color{blue}{-2}&\color{blue}{-4}&\color{orangered}{7} \end{array} $$The solution is:
$$ \frac{ x^{3}-6x^{2}+4x+23 }{ x-4 } = \color{blue}{x^{2}-2x-4} ~+~ \frac{ \color{red}{ 7 } }{ x-4 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&-6&4&23\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 1 }&-6&4&23\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&-6&4&23\\& & \color{blue}{4} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 4 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}4&1&\color{orangered}{ -6 }&4&23\\& & \color{orangered}{4} & & \\ \hline &1&\color{orangered}{-2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&-6&4&23\\& & 4& \color{blue}{-8} & \\ \hline &1&\color{blue}{-2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}4&1&-6&\color{orangered}{ 4 }&23\\& & 4& \color{orangered}{-8} & \\ \hline &1&-2&\color{orangered}{-4}& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&-6&4&23\\& & 4& -8& \color{blue}{-16} \\ \hline &1&-2&\color{blue}{-4}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 23 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrr}4&1&-6&4&\color{orangered}{ 23 }\\& & 4& -8& \color{orangered}{-16} \\ \hline &\color{blue}{1}&\color{blue}{-2}&\color{blue}{-4}&\color{orangered}{7} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-2x-4 } $ with a remainder of $ \color{red}{ 7 } $.