Divide using synthetic division $$ ( 2x^{4}-5x^{3}-x^{2}+3x+3 ) \div ( x-0.5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}\frac{ 1 }{ 2 }&2&-5&-1&3&3\\& & 1& -2& -\frac{ 3 }{ 2 }& \color{black}{\frac{ 3 }{ 4 }} \\ \hline &\color{blue}{2}&\color{blue}{-4}&\color{blue}{-3}&\color{blue}{\frac{ 3 }{ 2 }}&\color{orangered}{\frac{ 15 }{ 4 }} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-5x^{3}-x^{2}+3x+3 }{ x-\frac{ 1 }{ 2 } } = \color{blue}{2x^{3}-4x^{2}-3x+\frac{ 3 }{ 2 }} ~+~ \frac{ \color{red}{ \frac{ 15 }{ 4 } } }{ x-\frac{ 1 }{ 2 } } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{\frac{ 1 }{ 2 }}&2&-5&-1&3&3\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}\frac{ 1 }{ 2 }&\color{orangered}{ 2 }&-5&-1&3&3\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$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}{ 2 } = \color{blue}{ 1 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{\frac{ 1 }{ 2 }}&2&-5&-1&3&3\\& & \color{blue}{1} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 1 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}\frac{ 1 }{ 2 }&2&\color{orangered}{ -5 }&-1&3&3\\& & \color{orangered}{1} & & & \\ \hline &2&\color{orangered}{-4}&&& \end{array} $$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}{ \left( -4 \right) } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{\frac{ 1 }{ 2 }}&2&-5&-1&3&3\\& & 1& \color{blue}{-2} & & \\ \hline &2&\color{blue}{-4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}\frac{ 1 }{ 2 }&2&-5&\color{orangered}{ -1 }&3&3\\& & 1& \color{orangered}{-2} & & \\ \hline &2&-4&\color{orangered}{-3}&& \end{array} $$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}{ \left( -3 \right) } = \color{blue}{ -\frac{ 3 }{ 2 } } $.
$$ \begin{array}{c|rrrrr}\color{blue}{\frac{ 1 }{ 2 }}&2&-5&-1&3&3\\& & 1& -2& \color{blue}{-\frac{ 3 }{ 2 }} & \\ \hline &2&-4&\color{blue}{-3}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -\frac{ 3 }{ 2 } \right) } = \color{orangered}{ \frac{ 3 }{ 2 } } $
$$ \begin{array}{c|rrrrr}\frac{ 1 }{ 2 }&2&-5&-1&\color{orangered}{ 3 }&3\\& & 1& -2& \color{orangered}{-\frac{ 3 }{ 2 }} & \\ \hline &2&-4&-3&\color{orangered}{\frac{ 3 }{ 2 }}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ \frac{ 1 }{ 2 } } \cdot \color{blue}{ \frac{ 3 }{ 2 } } = \color{blue}{ \frac{ 3 }{ 4 } } $.
$$ \begin{array}{c|rrrrr}\color{blue}{\frac{ 1 }{ 2 }}&2&-5&-1&3&3\\& & 1& -2& -\frac{ 3 }{ 2 }& \color{blue}{\frac{ 3 }{ 4 }} \\ \hline &2&-4&-3&\color{blue}{\frac{ 3 }{ 2 }}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \frac{ 3 }{ 4 } } = \color{orangered}{ \frac{ 15 }{ 4 } } $
$$ \begin{array}{c|rrrrr}\frac{ 1 }{ 2 }&2&-5&-1&3&\color{orangered}{ 3 }\\& & 1& -2& -\frac{ 3 }{ 2 }& \color{orangered}{\frac{ 3 }{ 4 }} \\ \hline &\color{blue}{2}&\color{blue}{-4}&\color{blue}{-3}&\color{blue}{\frac{ 3 }{ 2 }}&\color{orangered}{\frac{ 15 }{ 4 }} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-4x^{2}-3x+\frac{ 3 }{ 2 } } $ with a remainder of $ \color{red}{ \frac{ 15 }{ 4 } } $.