Divide using synthetic division $$ ( 2x^{3}-8x^{2}+11x-6 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&2&-8&11&-6\\& & 4& -8& \color{black}{6} \\ \hline &\color{blue}{2}&\color{blue}{-4}&\color{blue}{3}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 2x^{3}-8x^{2}+11x-6 }{ x-2 } = \color{blue}{2x^{2}-4x+3} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{2}&2&-8&11&-6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 2 }&-8&11&-6\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 2 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&2&-8&11&-6\\& & \color{blue}{4} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 4 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}2&2&\color{orangered}{ -8 }&11&-6\\& & \color{orangered}{4} & & \\ \hline &2&\color{orangered}{-4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&2&-8&11&-6\\& & 4& \color{blue}{-8} & \\ \hline &2&\color{blue}{-4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}2&2&-8&\color{orangered}{ 11 }&-6\\& & 4& \color{orangered}{-8} & \\ \hline &2&-4&\color{orangered}{3}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 3 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&2&-8&11&-6\\& & 4& -8& \color{blue}{6} \\ \hline &2&-4&\color{blue}{3}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 6 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}2&2&-8&11&\color{orangered}{ -6 }\\& & 4& -8& \color{orangered}{6} \\ \hline &\color{blue}{2}&\color{blue}{-4}&\color{blue}{3}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-4x+3 } $ with a remainder of $ \color{red}{ 0 } $.