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