Divide using synthetic division $$ ( x^{3}+7x^{2}-12x+14 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&1&7&-12&14\\& & 1& 8& \color{black}{-4} \\ \hline &\color{blue}{1}&\color{blue}{8}&\color{blue}{-4}&\color{orangered}{10} \end{array} $$The solution is:
$$ \frac{ x^{3}+7x^{2}-12x+14 }{ x-1 } = \color{blue}{x^{2}+8x-4} ~+~ \frac{ \color{red}{ 10 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{1}&1&7&-12&14\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 1 }&7&-12&14\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&1&7&-12&14\\& & \color{blue}{1} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 1 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrr}1&1&\color{orangered}{ 7 }&-12&14\\& & \color{orangered}{1} & & \\ \hline &1&\color{orangered}{8}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 8 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&1&7&-12&14\\& & 1& \color{blue}{8} & \\ \hline &1&\color{blue}{8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 8 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}1&1&7&\color{orangered}{ -12 }&14\\& & 1& \color{orangered}{8} & \\ \hline &1&8&\color{orangered}{-4}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&1&7&-12&14\\& & 1& 8& \color{blue}{-4} \\ \hline &1&8&\color{blue}{-4}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 14 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrr}1&1&7&-12&\color{orangered}{ 14 }\\& & 1& 8& \color{orangered}{-4} \\ \hline &\color{blue}{1}&\color{blue}{8}&\color{blue}{-4}&\color{orangered}{10} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+8x-4 } $ with a remainder of $ \color{red}{ 10 } $.