Divide using synthetic division $$ ( 2x^{3}+1.5x+4.5 ) \div ( x+0.5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-\frac{ 1 }{ 2 }&2&0&\frac{ 3 }{ 2 }&\frac{ 9 }{ 2 }\\& & -1& \frac{ 1 }{ 2 }& \color{black}{-1} \\ \hline &\color{blue}{2}&\color{blue}{-1}&\color{blue}{2}&\color{orangered}{\frac{ 7 }{ 2 }} \end{array} $$The solution is:
$$ \frac{ 2x^{3}+\frac{ 3 }{ 2 }x+\frac{ 9 }{ 2 } }{ x+\frac{ 1 }{ 2 } } = \color{blue}{2x^{2}-x+2} ~+~ \frac{ \color{red}{ \frac{ 7 }{ 2 } } }{ x+\frac{ 1 }{ 2 } } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-\frac{ 1 }{ 2 }}&2&0&\frac{ 3 }{ 2 }&\frac{ 9 }{ 2 }\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-\frac{ 1 }{ 2 }&\color{orangered}{ 2 }&0&\frac{ 3 }{ 2 }&\frac{ 9 }{ 2 }\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-\frac{ 1 }{ 2 }}&2&0&\frac{ 3 }{ 2 }&\frac{ 9 }{ 2 }\\& & \color{blue}{-1} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -1 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}-\frac{ 1 }{ 2 }&2&\color{orangered}{ 0 }&\frac{ 3 }{ 2 }&\frac{ 9 }{ 2 }\\& & \color{orangered}{-1} & & \\ \hline &2&\color{orangered}{-1}&& \end{array} $$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( -1 \right) } = \color{blue}{ \frac{ 1 }{ 2 } } $.
$$ \begin{array}{c|rrrr}\color{blue}{-\frac{ 1 }{ 2 }}&2&0&\frac{ 3 }{ 2 }&\frac{ 9 }{ 2 }\\& & -1& \color{blue}{\frac{ 1 }{ 2 }} & \\ \hline &2&\color{blue}{-1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ \frac{ 3 }{ 2 } } + \color{orangered}{ \frac{ 1 }{ 2 } } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}-\frac{ 1 }{ 2 }&2&0&\color{orangered}{ \frac{ 3 }{ 2 } }&\frac{ 9 }{ 2 }\\& & -1& \color{orangered}{\frac{ 1 }{ 2 }} & \\ \hline &2&-1&\color{orangered}{2}& \end{array} $$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}{ 2 } = \color{blue}{ -1 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-\frac{ 1 }{ 2 }}&2&0&\frac{ 3 }{ 2 }&\frac{ 9 }{ 2 }\\& & -1& \frac{ 1 }{ 2 }& \color{blue}{-1} \\ \hline &2&-1&\color{blue}{2}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ \frac{ 9 }{ 2 } } + \color{orangered}{ \left( -1 \right) } = \color{orangered}{ \frac{ 7 }{ 2 } } $
$$ \begin{array}{c|rrrr}-\frac{ 1 }{ 2 }&2&0&\frac{ 3 }{ 2 }&\color{orangered}{ \frac{ 9 }{ 2 } }\\& & -1& \frac{ 1 }{ 2 }& \color{orangered}{-1} \\ \hline &\color{blue}{2}&\color{blue}{-1}&\color{blue}{2}&\color{orangered}{\frac{ 7 }{ 2 }} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-x+2 } $ with a remainder of $ \color{red}{ \frac{ 7 }{ 2 } } $.