Divide using synthetic division $$ ( 3x^{4}-2x^{3}-7x^{2}+4x+2 ) \div ( x+0.5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-\frac{ 1 }{ 2 }&3&-2&-7&4&2\\& & -\frac{ 3 }{ 2 }& \frac{ 7 }{ 4 }& \frac{ 21 }{ 8 }& \color{black}{-\frac{ 53 }{ 16 }} \\ \hline &\color{blue}{3}&\color{blue}{-\frac{ 7 }{ 2 }}&\color{blue}{-\frac{ 21 }{ 4 }}&\color{blue}{\frac{ 53 }{ 8 }}&\color{orangered}{-\frac{ 21 }{ 16 }} \end{array} $$The solution is:
$$ \frac{ 3x^{4}-2x^{3}-7x^{2}+4x+2 }{ x+\frac{ 1 }{ 2 } } = \color{blue}{3x^{3}-\frac{ 7 }{ 2 }x^{2}-\frac{ 21 }{ 4 }x+\frac{ 53 }{ 8 }} \color{red}{~-~} \frac{ \color{red}{ \frac{ 21 }{ 16 } } }{ x+\frac{ 1 }{ 2 } } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-\frac{ 1 }{ 2 }}&3&-2&-7&4&2\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-\frac{ 1 }{ 2 }&\color{orangered}{ 3 }&-2&-7&4&2\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -\frac{ 1 }{ 2 } } \cdot \color{blue}{ 3 } = \color{blue}{ -\frac{ 3 }{ 2 } } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-\frac{ 1 }{ 2 }}&3&-2&-7&4&2\\& & \color{blue}{-\frac{ 3 }{ 2 }} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -\frac{ 3 }{ 2 } \right) } = \color{orangered}{ -\frac{ 7 }{ 2 } } $
$$ \begin{array}{c|rrrrr}-\frac{ 1 }{ 2 }&3&\color{orangered}{ -2 }&-7&4&2\\& & \color{orangered}{-\frac{ 3 }{ 2 }} & & & \\ \hline &3&\color{orangered}{-\frac{ 7 }{ 2 }}&&& \end{array} $$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( -\frac{ 7 }{ 2 } \right) } = \color{blue}{ \frac{ 7 }{ 4 } } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-\frac{ 1 }{ 2 }}&3&-2&-7&4&2\\& & -\frac{ 3 }{ 2 }& \color{blue}{\frac{ 7 }{ 4 }} & & \\ \hline &3&\color{blue}{-\frac{ 7 }{ 2 }}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ \frac{ 7 }{ 4 } } = \color{orangered}{ -\frac{ 21 }{ 4 } } $
$$ \begin{array}{c|rrrrr}-\frac{ 1 }{ 2 }&3&-2&\color{orangered}{ -7 }&4&2\\& & -\frac{ 3 }{ 2 }& \color{orangered}{\frac{ 7 }{ 4 }} & & \\ \hline &3&-\frac{ 7 }{ 2 }&\color{orangered}{-\frac{ 21 }{ 4 }}&& \end{array} $$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( -\frac{ 21 }{ 4 } \right) } = \color{blue}{ \frac{ 21 }{ 8 } } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-\frac{ 1 }{ 2 }}&3&-2&-7&4&2\\& & -\frac{ 3 }{ 2 }& \frac{ 7 }{ 4 }& \color{blue}{\frac{ 21 }{ 8 }} & \\ \hline &3&-\frac{ 7 }{ 2 }&\color{blue}{-\frac{ 21 }{ 4 }}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \frac{ 21 }{ 8 } } = \color{orangered}{ \frac{ 53 }{ 8 } } $
$$ \begin{array}{c|rrrrr}-\frac{ 1 }{ 2 }&3&-2&-7&\color{orangered}{ 4 }&2\\& & -\frac{ 3 }{ 2 }& \frac{ 7 }{ 4 }& \color{orangered}{\frac{ 21 }{ 8 }} & \\ \hline &3&-\frac{ 7 }{ 2 }&-\frac{ 21 }{ 4 }&\color{orangered}{\frac{ 53 }{ 8 }}& \end{array} $$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{ 53 }{ 8 } } = \color{blue}{ -\frac{ 53 }{ 16 } } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-\frac{ 1 }{ 2 }}&3&-2&-7&4&2\\& & -\frac{ 3 }{ 2 }& \frac{ 7 }{ 4 }& \frac{ 21 }{ 8 }& \color{blue}{-\frac{ 53 }{ 16 }} \\ \hline &3&-\frac{ 7 }{ 2 }&-\frac{ 21 }{ 4 }&\color{blue}{\frac{ 53 }{ 8 }}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -\frac{ 53 }{ 16 } \right) } = \color{orangered}{ -\frac{ 21 }{ 16 } } $
$$ \begin{array}{c|rrrrr}-\frac{ 1 }{ 2 }&3&-2&-7&4&\color{orangered}{ 2 }\\& & -\frac{ 3 }{ 2 }& \frac{ 7 }{ 4 }& \frac{ 21 }{ 8 }& \color{orangered}{-\frac{ 53 }{ 16 }} \\ \hline &\color{blue}{3}&\color{blue}{-\frac{ 7 }{ 2 }}&\color{blue}{-\frac{ 21 }{ 4 }}&\color{blue}{\frac{ 53 }{ 8 }}&\color{orangered}{-\frac{ 21 }{ 16 }} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}-\frac{ 7 }{ 2 }x^{2}-\frac{ 21 }{ 4 }x+\frac{ 53 }{ 8 } } $ with a remainder of $ \color{red}{ -\frac{ 21 }{ 16 } } $.