Divide using synthetic division $$ ( x^{3}-5x^{2}+4x+6 ) \div ( x-0.5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}\frac{ 1 }{ 2 }&1&-5&4&6\\& & \frac{ 1 }{ 2 }& -\frac{ 9 }{ 4 }& \color{black}{\frac{ 7 }{ 8 }} \\ \hline &\color{blue}{1}&\color{blue}{-\frac{ 9 }{ 2 }}&\color{blue}{\frac{ 7 }{ 4 }}&\color{orangered}{\frac{ 55 }{ 8 }} \end{array} $$The solution is:
$$ \frac{ x^{3}-5x^{2}+4x+6 }{ x-\frac{ 1 }{ 2 } } = \color{blue}{x^{2}-\frac{ 9 }{ 2 }x+\frac{ 7 }{ 4 }} ~+~ \frac{ \color{red}{ \frac{ 55 }{ 8 } } }{ x-\frac{ 1 }{ 2 } } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -\frac{ 1 }{ 2 } = 0 $ ( $ x = \color{blue}{ \frac{ 1 }{ 2 } } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{\frac{ 1 }{ 2 }}&1&-5&4&6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}\frac{ 1 }{ 2 }&\color{orangered}{ 1 }&-5&4&6\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ \frac{ 1 }{ 2 } } \cdot \color{blue}{ 1 } = \color{blue}{ \frac{ 1 }{ 2 } } $.
$$ \begin{array}{c|rrrr}\color{blue}{\frac{ 1 }{ 2 }}&1&-5&4&6\\& & \color{blue}{\frac{ 1 }{ 2 }} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ \frac{ 1 }{ 2 } } = \color{orangered}{ -\frac{ 9 }{ 2 } } $
$$ \begin{array}{c|rrrr}\frac{ 1 }{ 2 }&1&\color{orangered}{ -5 }&4&6\\& & \color{orangered}{\frac{ 1 }{ 2 }} & & \\ \hline &1&\color{orangered}{-\frac{ 9 }{ 2 }}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ \frac{ 1 }{ 2 } } \cdot \color{blue}{ \left( -\frac{ 9 }{ 2 } \right) } = \color{blue}{ -\frac{ 9 }{ 4 } } $.
$$ \begin{array}{c|rrrr}\color{blue}{\frac{ 1 }{ 2 }}&1&-5&4&6\\& & \frac{ 1 }{ 2 }& \color{blue}{-\frac{ 9 }{ 4 }} & \\ \hline &1&\color{blue}{-\frac{ 9 }{ 2 }}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -\frac{ 9 }{ 4 } \right) } = \color{orangered}{ \frac{ 7 }{ 4 } } $
$$ \begin{array}{c|rrrr}\frac{ 1 }{ 2 }&1&-5&\color{orangered}{ 4 }&6\\& & \frac{ 1 }{ 2 }& \color{orangered}{-\frac{ 9 }{ 4 }} & \\ \hline &1&-\frac{ 9 }{ 2 }&\color{orangered}{\frac{ 7 }{ 4 }}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ \frac{ 1 }{ 2 } } \cdot \color{blue}{ \frac{ 7 }{ 4 } } = \color{blue}{ \frac{ 7 }{ 8 } } $.
$$ \begin{array}{c|rrrr}\color{blue}{\frac{ 1 }{ 2 }}&1&-5&4&6\\& & \frac{ 1 }{ 2 }& -\frac{ 9 }{ 4 }& \color{blue}{\frac{ 7 }{ 8 }} \\ \hline &1&-\frac{ 9 }{ 2 }&\color{blue}{\frac{ 7 }{ 4 }}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \frac{ 7 }{ 8 } } = \color{orangered}{ \frac{ 55 }{ 8 } } $
$$ \begin{array}{c|rrrr}\frac{ 1 }{ 2 }&1&-5&4&\color{orangered}{ 6 }\\& & \frac{ 1 }{ 2 }& -\frac{ 9 }{ 4 }& \color{orangered}{\frac{ 7 }{ 8 }} \\ \hline &\color{blue}{1}&\color{blue}{-\frac{ 9 }{ 2 }}&\color{blue}{\frac{ 7 }{ 4 }}&\color{orangered}{\frac{ 55 }{ 8 }} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-\frac{ 9 }{ 2 }x+\frac{ 7 }{ 4 } } $ with a remainder of $ \color{red}{ \frac{ 55 }{ 8 } } $.