Divide using synthetic division $$ ( 2x^{4}-9x^{3}+9x^{2}+5x-1 ) \div ( 2x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-1&2&-9&9&5&-1\\& & -2& 11& -20& \color{black}{15} \\ \hline &\color{blue}{2}&\color{blue}{-11}&\color{blue}{20}&\color{blue}{-15}&\color{orangered}{14} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-9x^{3}+9x^{2}+5x-1 }{ x+1 } = \color{blue}{2x^{3}-11x^{2}+20x-15} ~+~ \frac{ \color{red}{ 14 } }{ x+1 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 1 = 0 $ ( $ x = \color{blue}{ -1 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&2&-9&9&5&-1\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-1&\color{orangered}{ 2 }&-9&9&5&-1\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 2 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&2&-9&9&5&-1\\& & \color{blue}{-2} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -11 } $
$$ \begin{array}{c|rrrrr}-1&2&\color{orangered}{ -9 }&9&5&-1\\& & \color{orangered}{-2} & & & \\ \hline &2&\color{orangered}{-11}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -11 \right) } = \color{blue}{ 11 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&2&-9&9&5&-1\\& & -2& \color{blue}{11} & & \\ \hline &2&\color{blue}{-11}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ 11 } = \color{orangered}{ 20 } $
$$ \begin{array}{c|rrrrr}-1&2&-9&\color{orangered}{ 9 }&5&-1\\& & -2& \color{orangered}{11} & & \\ \hline &2&-11&\color{orangered}{20}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 20 } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&2&-9&9&5&-1\\& & -2& 11& \color{blue}{-20} & \\ \hline &2&-11&\color{blue}{20}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -15 } $
$$ \begin{array}{c|rrrrr}-1&2&-9&9&\color{orangered}{ 5 }&-1\\& & -2& 11& \color{orangered}{-20} & \\ \hline &2&-11&20&\color{orangered}{-15}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -15 \right) } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&2&-9&9&5&-1\\& & -2& 11& -20& \color{blue}{15} \\ \hline &2&-11&20&\color{blue}{-15}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 15 } = \color{orangered}{ 14 } $
$$ \begin{array}{c|rrrrr}-1&2&-9&9&5&\color{orangered}{ -1 }\\& & -2& 11& -20& \color{orangered}{15} \\ \hline &\color{blue}{2}&\color{blue}{-11}&\color{blue}{20}&\color{blue}{-15}&\color{orangered}{14} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-11x^{2}+20x-15 } $ with a remainder of $ \color{red}{ 14 } $.