Divide using synthetic division $$ ( 2x^{4}-5x^{3}-x^{2}+3x+3 ) \div ( x+0.5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-\frac{ 1 }{ 2 }&2&-5&-1&3&3\\& & -1& 3& -1& \color{black}{-1} \\ \hline &\color{blue}{2}&\color{blue}{-6}&\color{blue}{2}&\color{blue}{2}&\color{orangered}{2} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-5x^{3}-x^{2}+3x+3 }{ x+\frac{ 1 }{ 2 } } = \color{blue}{2x^{3}-6x^{2}+2x+2} ~+~ \frac{ \color{red}{ 2 } }{ x+\frac{ 1 }{ 2 } } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-\frac{ 1 }{ 2 }}&2&-5&-1&3&3\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-\frac{ 1 }{ 2 }&\color{orangered}{ 2 }&-5&-1&3&3\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-\frac{ 1 }{ 2 }}&2&-5&-1&3&3\\& & \color{blue}{-1} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ \left( -1 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}-\frac{ 1 }{ 2 }&2&\color{orangered}{ -5 }&-1&3&3\\& & \color{orangered}{-1} & & & \\ \hline &2&\color{orangered}{-6}&&& \end{array} $$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( -6 \right) } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-\frac{ 1 }{ 2 }}&2&-5&-1&3&3\\& & -1& \color{blue}{3} & & \\ \hline &2&\color{blue}{-6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 3 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}-\frac{ 1 }{ 2 }&2&-5&\color{orangered}{ -1 }&3&3\\& & -1& \color{orangered}{3} & & \\ \hline &2&-6&\color{orangered}{2}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-\frac{ 1 }{ 2 }}&2&-5&-1&3&3\\& & -1& 3& \color{blue}{-1} & \\ \hline &2&-6&\color{blue}{2}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -1 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}-\frac{ 1 }{ 2 }&2&-5&-1&\color{orangered}{ 3 }&3\\& & -1& 3& \color{orangered}{-1} & \\ \hline &2&-6&2&\color{orangered}{2}& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-\frac{ 1 }{ 2 }}&2&-5&-1&3&3\\& & -1& 3& -1& \color{blue}{-1} \\ \hline &2&-6&2&\color{blue}{2}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -1 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}-\frac{ 1 }{ 2 }&2&-5&-1&3&\color{orangered}{ 3 }\\& & -1& 3& -1& \color{orangered}{-1} \\ \hline &\color{blue}{2}&\color{blue}{-6}&\color{blue}{2}&\color{blue}{2}&\color{orangered}{2} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-6x^{2}+2x+2 } $ with a remainder of $ \color{red}{ 2 } $.