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