Divide using synthetic division $$ ( 2x^{4}-x^{3}+2x-1 ) \div ( x-0.5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}\frac{ 1 }{ 2 }&2&-1&0&2&-1\\& & 1& 0& 0& \color{black}{1} \\ \hline &\color{blue}{2}&\color{blue}{0}&\color{blue}{0}&\color{blue}{2}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-x^{3}+2x-1 }{ x-\frac{ 1 }{ 2 } } = \color{blue}{2x^{3}+2} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{\frac{ 1 }{ 2 }}&2&-1&0&2&-1\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}\frac{ 1 }{ 2 }&\color{orangered}{ 2 }&-1&0&2&-1\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{\frac{ 1 }{ 2 }}&2&-1&0&2&-1\\& & \color{blue}{1} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 1 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}\frac{ 1 }{ 2 }&2&\color{orangered}{ -1 }&0&2&-1\\& & \color{orangered}{1} & & & \\ \hline &2&\color{orangered}{0}&&& \end{array} $$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}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{\frac{ 1 }{ 2 }}&2&-1&0&2&-1\\& & 1& \color{blue}{0} & & \\ \hline &2&\color{blue}{0}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}\frac{ 1 }{ 2 }&2&-1&\color{orangered}{ 0 }&2&-1\\& & 1& \color{orangered}{0} & & \\ \hline &2&0&\color{orangered}{0}&& \end{array} $$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}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{\frac{ 1 }{ 2 }}&2&-1&0&2&-1\\& & 1& 0& \color{blue}{0} & \\ \hline &2&0&\color{blue}{0}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 0 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}\frac{ 1 }{ 2 }&2&-1&0&\color{orangered}{ 2 }&-1\\& & 1& 0& \color{orangered}{0} & \\ \hline &2&0&0&\color{orangered}{2}& \end{array} $$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}{ 2 } = \color{blue}{ 1 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{\frac{ 1 }{ 2 }}&2&-1&0&2&-1\\& & 1& 0& 0& \color{blue}{1} \\ \hline &2&0&0&\color{blue}{2}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 1 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}\frac{ 1 }{ 2 }&2&-1&0&2&\color{orangered}{ -1 }\\& & 1& 0& 0& \color{orangered}{1} \\ \hline &\color{blue}{2}&\color{blue}{0}&\color{blue}{0}&\color{blue}{2}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+2 } $ with a remainder of $ \color{red}{ 0 } $.