Divide using synthetic division $$ ( 4x^{2}+2x+3 ) \div ( x-0.5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrr}\frac{ 1 }{ 2 }&4&2&3\\& & 2& \color{black}{2} \\ \hline &\color{blue}{4}&\color{blue}{4}&\color{orangered}{5} \end{array} $$The solution is:
$$ \frac{ 4x^{2}+2x+3 }{ x-\frac{ 1 }{ 2 } } = \color{blue}{4x+4} ~+~ \frac{ \color{red}{ 5 } }{ x-\frac{ 1 }{ 2 } } $$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.
$$ \begin{array}{c|rrr}\color{blue}{\frac{ 1 }{ 2 }}&4&2&3\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}\frac{ 1 }{ 2 }&\color{orangered}{ 4 }&2&3\\& & & \\ \hline &\color{orangered}{4}&& \end{array} $$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}{ 4 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrr}\color{blue}{\frac{ 1 }{ 2 }}&4&2&3\\& & \color{blue}{2} & \\ \hline &\color{blue}{4}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 2 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrr}\frac{ 1 }{ 2 }&4&\color{orangered}{ 2 }&3\\& & \color{orangered}{2} & \\ \hline &4&\color{orangered}{4}& \end{array} $$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}{ 4 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrr}\color{blue}{\frac{ 1 }{ 2 }}&4&2&3\\& & 2& \color{blue}{2} \\ \hline &4&\color{blue}{4}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 2 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrr}\frac{ 1 }{ 2 }&4&2&\color{orangered}{ 3 }\\& & 2& \color{orangered}{2} \\ \hline &\color{blue}{4}&\color{blue}{4}&\color{orangered}{5} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x+4 } $ with a remainder of $ \color{red}{ 5 } $.