Divide using synthetic division $$ ( 2x^{3}-x^{2}-16x-9 ) \div ( x+0.5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-\frac{ 1 }{ 2 }&2&-1&-16&-9\\& & -1& 1& \color{black}{\frac{ 15 }{ 2 }} \\ \hline &\color{blue}{2}&\color{blue}{-2}&\color{blue}{-15}&\color{orangered}{-\frac{ 3 }{ 2 }} \end{array} $$The solution is:
$$ \frac{ 2x^{3}-x^{2}-16x-9 }{ x+\frac{ 1 }{ 2 } } = \color{blue}{2x^{2}-2x-15} \color{red}{~-~} \frac{ \color{red}{ \frac{ 3 }{ 2 } } }{ x+\frac{ 1 }{ 2 } } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-\frac{ 1 }{ 2 }}&2&-1&-16&-9\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-\frac{ 1 }{ 2 }&\color{orangered}{ 2 }&-1&-16&-9\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-\frac{ 1 }{ 2 }}&2&-1&-16&-9\\& & \color{blue}{-1} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -1 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}-\frac{ 1 }{ 2 }&2&\color{orangered}{ -1 }&-16&-9\\& & \color{orangered}{-1} & & \\ \hline &2&\color{orangered}{-2}&& \end{array} $$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( -2 \right) } = \color{blue}{ 1 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-\frac{ 1 }{ 2 }}&2&-1&-16&-9\\& & -1& \color{blue}{1} & \\ \hline &2&\color{blue}{-2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 1 } = \color{orangered}{ -15 } $
$$ \begin{array}{c|rrrr}-\frac{ 1 }{ 2 }&2&-1&\color{orangered}{ -16 }&-9\\& & -1& \color{orangered}{1} & \\ \hline &2&-2&\color{orangered}{-15}& \end{array} $$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( -15 \right) } = \color{blue}{ \frac{ 15 }{ 2 } } $.
$$ \begin{array}{c|rrrr}\color{blue}{-\frac{ 1 }{ 2 }}&2&-1&-16&-9\\& & -1& 1& \color{blue}{\frac{ 15 }{ 2 }} \\ \hline &2&-2&\color{blue}{-15}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ \frac{ 15 }{ 2 } } = \color{orangered}{ -\frac{ 3 }{ 2 } } $
$$ \begin{array}{c|rrrr}-\frac{ 1 }{ 2 }&2&-1&-16&\color{orangered}{ -9 }\\& & -1& 1& \color{orangered}{\frac{ 15 }{ 2 }} \\ \hline &\color{blue}{2}&\color{blue}{-2}&\color{blue}{-15}&\color{orangered}{-\frac{ 3 }{ 2 }} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-2x-15 } $ with a remainder of $ \color{red}{ -\frac{ 3 }{ 2 } } $.