Divide using synthetic division $$ ( 2x^{4}-9x^{3}+9x^{2}+5x-1 ) \div ( 2x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&2&-9&9&5&-1\\& & 2& -7& 2& \color{black}{7} \\ \hline &\color{blue}{2}&\color{blue}{-7}&\color{blue}{2}&\color{blue}{7}&\color{orangered}{6} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-9x^{3}+9x^{2}+5x-1 }{ x-1 } = \color{blue}{2x^{3}-7x^{2}+2x+7} ~+~ \frac{ \color{red}{ 6 } }{ x-1 } $$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}{ 2 } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrr}1&2&\color{orangered}{ -9 }&9&5&-1\\& & \color{orangered}{2} & & & \\ \hline &2&\color{orangered}{-7}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ -7 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&2&-9&9&5&-1\\& & 2& \color{blue}{-7} & & \\ \hline &2&\color{blue}{-7}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -7 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}1&2&-9&\color{orangered}{ 9 }&5&-1\\& & 2& \color{orangered}{-7} & & \\ \hline &2&-7&\color{orangered}{2}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 2 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&2&-9&9&5&-1\\& & 2& -7& \color{blue}{2} & \\ \hline &2&-7&\color{blue}{2}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 2 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrr}1&2&-9&9&\color{orangered}{ 5 }&-1\\& & 2& -7& \color{orangered}{2} & \\ \hline &2&-7&2&\color{orangered}{7}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 7 } = \color{blue}{ 7 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&2&-9&9&5&-1\\& & 2& -7& 2& \color{blue}{7} \\ \hline &2&-7&2&\color{blue}{7}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 7 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}1&2&-9&9&5&\color{orangered}{ -1 }\\& & 2& -7& 2& \color{orangered}{7} \\ \hline &\color{blue}{2}&\color{blue}{-7}&\color{blue}{2}&\color{blue}{7}&\color{orangered}{6} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-7x^{2}+2x+7 } $ with a remainder of $ \color{red}{ 6 } $.