Divide using synthetic division $$ ( 2x^{5}+x^{4}-10x^{3}-5x^{2}+10x+7 ) \div ( 2x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}1&2&1&-10&-5&10&7\\& & 2& 3& -7& -12& \color{black}{-2} \\ \hline &\color{blue}{2}&\color{blue}{3}&\color{blue}{-7}&\color{blue}{-12}&\color{blue}{-2}&\color{orangered}{5} \end{array} $$The solution is:
$$ \frac{ 2x^{5}+x^{4}-10x^{3}-5x^{2}+10x+7 }{ x-1 } = \color{blue}{2x^{4}+3x^{3}-7x^{2}-12x-2} ~+~ \frac{ \color{red}{ 5 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&2&1&-10&-5&10&7\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}1&\color{orangered}{ 2 }&1&-10&-5&10&7\\& & & & & & \\ \hline &\color{orangered}{2}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&2&1&-10&-5&10&7\\& & \color{blue}{2} & & & & \\ \hline &\color{blue}{2}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 2 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrrr}1&2&\color{orangered}{ 1 }&-10&-5&10&7\\& & \color{orangered}{2} & & & & \\ \hline &2&\color{orangered}{3}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 3 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&2&1&-10&-5&10&7\\& & 2& \color{blue}{3} & & & \\ \hline &2&\color{blue}{3}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 3 } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrrr}1&2&1&\color{orangered}{ -10 }&-5&10&7\\& & 2& \color{orangered}{3} & & & \\ \hline &2&3&\color{orangered}{-7}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ -7 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&2&1&-10&-5&10&7\\& & 2& 3& \color{blue}{-7} & & \\ \hline &2&3&\color{blue}{-7}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ \left( -7 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrrrr}1&2&1&-10&\color{orangered}{ -5 }&10&7\\& & 2& 3& \color{orangered}{-7} & & \\ \hline &2&3&-7&\color{orangered}{-12}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&2&1&-10&-5&10&7\\& & 2& 3& -7& \color{blue}{-12} & \\ \hline &2&3&-7&\color{blue}{-12}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrrr}1&2&1&-10&-5&\color{orangered}{ 10 }&7\\& & 2& 3& -7& \color{orangered}{-12} & \\ \hline &2&3&-7&-12&\color{orangered}{-2}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&2&1&-10&-5&10&7\\& & 2& 3& -7& -12& \color{blue}{-2} \\ \hline &2&3&-7&-12&\color{blue}{-2}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrrr}1&2&1&-10&-5&10&\color{orangered}{ 7 }\\& & 2& 3& -7& -12& \color{orangered}{-2} \\ \hline &\color{blue}{2}&\color{blue}{3}&\color{blue}{-7}&\color{blue}{-12}&\color{blue}{-2}&\color{orangered}{5} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{4}+3x^{3}-7x^{2}-12x-2 } $ with a remainder of $ \color{red}{ 5 } $.