Divide using synthetic division $$ ( x^{3}+6x^{2}+3x-10 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&1&6&3&-10\\& & 1& 7& \color{black}{10} \\ \hline &\color{blue}{1}&\color{blue}{7}&\color{blue}{10}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{3}+6x^{2}+3x-10 }{ x-1 } = \color{blue}{x^{2}+7x+10} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{1}&1&6&3&-10\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 1 }&6&3&-10\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 1 } = \color{blue}{ 1 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&1&6&3&-10\\& & \color{blue}{1} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 1 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrr}1&1&\color{orangered}{ 6 }&3&-10\\& & \color{orangered}{1} & & \\ \hline &1&\color{orangered}{7}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 7 } = \color{blue}{ 7 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&1&6&3&-10\\& & 1& \color{blue}{7} & \\ \hline &1&\color{blue}{7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 7 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrr}1&1&6&\color{orangered}{ 3 }&-10\\& & 1& \color{orangered}{7} & \\ \hline &1&7&\color{orangered}{10}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 10 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&1&6&3&-10\\& & 1& 7& \color{blue}{10} \\ \hline &1&7&\color{blue}{10}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 10 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}1&1&6&3&\color{orangered}{ -10 }\\& & 1& 7& \color{orangered}{10} \\ \hline &\color{blue}{1}&\color{blue}{7}&\color{blue}{10}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+7x+10 } $ with a remainder of $ \color{red}{ 0 } $.