Divide using synthetic division $$ ( 9x^{3}-73x^{2}+71x-2 ) \div ( x-7 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}7&9&-73&71&-2\\& & 63& -70& \color{black}{7} \\ \hline &\color{blue}{9}&\color{blue}{-10}&\color{blue}{1}&\color{orangered}{5} \end{array} $$The solution is:
$$ \frac{ 9x^{3}-73x^{2}+71x-2 }{ x-7 } = \color{blue}{9x^{2}-10x+1} ~+~ \frac{ \color{red}{ 5 } }{ x-7 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -7 = 0 $ ( $ x = \color{blue}{ 7 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{7}&9&-73&71&-2\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}7&\color{orangered}{ 9 }&-73&71&-2\\& & & & \\ \hline &\color{orangered}{9}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 9 } = \color{blue}{ 63 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&9&-73&71&-2\\& & \color{blue}{63} & & \\ \hline &\color{blue}{9}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -73 } + \color{orangered}{ 63 } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrr}7&9&\color{orangered}{ -73 }&71&-2\\& & \color{orangered}{63} & & \\ \hline &9&\color{orangered}{-10}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ -70 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&9&-73&71&-2\\& & 63& \color{blue}{-70} & \\ \hline &9&\color{blue}{-10}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 71 } + \color{orangered}{ \left( -70 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrr}7&9&-73&\color{orangered}{ 71 }&-2\\& & 63& \color{orangered}{-70} & \\ \hline &9&-10&\color{orangered}{1}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 1 } = \color{blue}{ 7 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&9&-73&71&-2\\& & 63& -70& \color{blue}{7} \\ \hline &9&-10&\color{blue}{1}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 7 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}7&9&-73&71&\color{orangered}{ -2 }\\& & 63& -70& \color{orangered}{7} \\ \hline &\color{blue}{9}&\color{blue}{-10}&\color{blue}{1}&\color{orangered}{5} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 9x^{2}-10x+1 } $ with a remainder of $ \color{red}{ 5 } $.