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