Divide using synthetic division $$ ( -3x^{3}-0.5x^{2}-2 ) \div ( x-1.5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}\frac{ 3 }{ 2 }&-3&-\frac{ 1 }{ 2 }&0&-2\\& & -\frac{ 9 }{ 2 }& -\frac{ 15 }{ 2 }& \color{black}{-\frac{ 45 }{ 4 }} \\ \hline &\color{blue}{-3}&\color{blue}{-5}&\color{blue}{-\frac{ 15 }{ 2 }}&\color{orangered}{-\frac{ 53 }{ 4 }} \end{array} $$The solution is:
$$ \frac{ -3x^{3}-\frac{ 1 }{ 2 }x^{2}-2 }{ x-\frac{ 3 }{ 2 } } = \color{blue}{-3x^{2}-5x-\frac{ 15 }{ 2 }} \color{red}{~-~} \frac{ \color{red}{ \frac{ 53 }{ 4 } } }{ x-\frac{ 3 }{ 2 } } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -\frac{ 3 }{ 2 } = 0 $ ( $ x = \color{blue}{ \frac{ 3 }{ 2 } } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{\frac{ 3 }{ 2 }}&-3&-\frac{ 1 }{ 2 }&0&-2\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}\frac{ 3 }{ 2 }&\color{orangered}{ -3 }&-\frac{ 1 }{ 2 }&0&-2\\& & & & \\ \hline &\color{orangered}{-3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ \frac{ 3 }{ 2 } } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -\frac{ 9 }{ 2 } } $.
$$ \begin{array}{c|rrrr}\color{blue}{\frac{ 3 }{ 2 }}&-3&-\frac{ 1 }{ 2 }&0&-2\\& & \color{blue}{-\frac{ 9 }{ 2 }} & & \\ \hline &\color{blue}{-3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -\frac{ 1 }{ 2 } } + \color{orangered}{ \left( -\frac{ 9 }{ 2 } \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}\frac{ 3 }{ 2 }&-3&\color{orangered}{ -\frac{ 1 }{ 2 } }&0&-2\\& & \color{orangered}{-\frac{ 9 }{ 2 }} & & \\ \hline &-3&\color{orangered}{-5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ \frac{ 3 }{ 2 } } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -\frac{ 15 }{ 2 } } $.
$$ \begin{array}{c|rrrr}\color{blue}{\frac{ 3 }{ 2 }}&-3&-\frac{ 1 }{ 2 }&0&-2\\& & -\frac{ 9 }{ 2 }& \color{blue}{-\frac{ 15 }{ 2 }} & \\ \hline &-3&\color{blue}{-5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -\frac{ 15 }{ 2 } \right) } = \color{orangered}{ -\frac{ 15 }{ 2 } } $
$$ \begin{array}{c|rrrr}\frac{ 3 }{ 2 }&-3&-\frac{ 1 }{ 2 }&\color{orangered}{ 0 }&-2\\& & -\frac{ 9 }{ 2 }& \color{orangered}{-\frac{ 15 }{ 2 }} & \\ \hline &-3&-5&\color{orangered}{-\frac{ 15 }{ 2 }}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ \frac{ 3 }{ 2 } } \cdot \color{blue}{ \left( -\frac{ 15 }{ 2 } \right) } = \color{blue}{ -\frac{ 45 }{ 4 } } $.
$$ \begin{array}{c|rrrr}\color{blue}{\frac{ 3 }{ 2 }}&-3&-\frac{ 1 }{ 2 }&0&-2\\& & -\frac{ 9 }{ 2 }& -\frac{ 15 }{ 2 }& \color{blue}{-\frac{ 45 }{ 4 }} \\ \hline &-3&-5&\color{blue}{-\frac{ 15 }{ 2 }}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -\frac{ 45 }{ 4 } \right) } = \color{orangered}{ -\frac{ 53 }{ 4 } } $
$$ \begin{array}{c|rrrr}\frac{ 3 }{ 2 }&-3&-\frac{ 1 }{ 2 }&0&\color{orangered}{ -2 }\\& & -\frac{ 9 }{ 2 }& -\frac{ 15 }{ 2 }& \color{orangered}{-\frac{ 45 }{ 4 }} \\ \hline &\color{blue}{-3}&\color{blue}{-5}&\color{blue}{-\frac{ 15 }{ 2 }}&\color{orangered}{-\frac{ 53 }{ 4 }} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -3x^{2}-5x-\frac{ 15 }{ 2 } } $ with a remainder of $ \color{red}{ -\frac{ 53 }{ 4 } } $.