Divide using synthetic division $$ ( 2x^{3}-13x^{2}-8x+7 ) \div ( 2x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&2&-13&-8&7\\& & 2& -11& \color{black}{-19} \\ \hline &\color{blue}{2}&\color{blue}{-11}&\color{blue}{-19}&\color{orangered}{-12} \end{array} $$The solution is:
$$ \frac{ 2x^{3}-13x^{2}-8x+7 }{ x-1 } = \color{blue}{2x^{2}-11x-19} \color{red}{~-~} \frac{ \color{red}{ 12 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{1}&2&-13&-8&7\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 2 }&-13&-8&7\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&2&-13&-8&7\\& & \color{blue}{2} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -13 } + \color{orangered}{ 2 } = \color{orangered}{ -11 } $
$$ \begin{array}{c|rrrr}1&2&\color{orangered}{ -13 }&-8&7\\& & \color{orangered}{2} & & \\ \hline &2&\color{orangered}{-11}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&2&-13&-8&7\\& & 2& \color{blue}{-11} & \\ \hline &2&\color{blue}{-11}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ \left( -11 \right) } = \color{orangered}{ -19 } $
$$ \begin{array}{c|rrrr}1&2&-13&\color{orangered}{ -8 }&7\\& & 2& \color{orangered}{-11} & \\ \hline &2&-11&\color{orangered}{-19}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -19 \right) } = \color{blue}{ -19 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&2&-13&-8&7\\& & 2& -11& \color{blue}{-19} \\ \hline &2&-11&\color{blue}{-19}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -19 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrr}1&2&-13&-8&\color{orangered}{ 7 }\\& & 2& -11& \color{orangered}{-19} \\ \hline &\color{blue}{2}&\color{blue}{-11}&\color{blue}{-19}&\color{orangered}{-12} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-11x-19 } $ with a remainder of $ \color{red}{ -12 } $.