Divide using synthetic division $$ ( 7x^{3}-2x^{2}+4x-6 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&7&-2&4&-6\\& & 7& 5& \color{black}{9} \\ \hline &\color{blue}{7}&\color{blue}{5}&\color{blue}{9}&\color{orangered}{3} \end{array} $$The solution is:
$$ \frac{ 7x^{3}-2x^{2}+4x-6 }{ x-1 } = \color{blue}{7x^{2}+5x+9} ~+~ \frac{ \color{red}{ 3 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{1}&7&-2&4&-6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 7 }&-2&4&-6\\& & & & \\ \hline &\color{orangered}{7}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 7 } = \color{blue}{ 7 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&7&-2&4&-6\\& & \color{blue}{7} & & \\ \hline &\color{blue}{7}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 7 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}1&7&\color{orangered}{ -2 }&4&-6\\& & \color{orangered}{7} & & \\ \hline &7&\color{orangered}{5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 5 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&7&-2&4&-6\\& & 7& \color{blue}{5} & \\ \hline &7&\color{blue}{5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 5 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrr}1&7&-2&\color{orangered}{ 4 }&-6\\& & 7& \color{orangered}{5} & \\ \hline &7&5&\color{orangered}{9}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 9 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&7&-2&4&-6\\& & 7& 5& \color{blue}{9} \\ \hline &7&5&\color{blue}{9}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 9 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}1&7&-2&4&\color{orangered}{ -6 }\\& & 7& 5& \color{orangered}{9} \\ \hline &\color{blue}{7}&\color{blue}{5}&\color{blue}{9}&\color{orangered}{3} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 7x^{2}+5x+9 } $ with a remainder of $ \color{red}{ 3 } $.