Divide using synthetic division $$ ( x^{3}+4x^{2}-7x+1 ) \div ( x-7 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}7&1&4&-7&1\\& & 7& 77& \color{black}{490} \\ \hline &\color{blue}{1}&\color{blue}{11}&\color{blue}{70}&\color{orangered}{491} \end{array} $$The solution is:
$$ \frac{ x^{3}+4x^{2}-7x+1 }{ x-7 } = \color{blue}{x^{2}+11x+70} ~+~ \frac{ \color{red}{ 491 } }{ x-7 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{7}&1&4&-7&1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}7&\color{orangered}{ 1 }&4&-7&1\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 1 } = \color{blue}{ 7 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&1&4&-7&1\\& & \color{blue}{7} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 7 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrr}7&1&\color{orangered}{ 4 }&-7&1\\& & \color{orangered}{7} & & \\ \hline &1&\color{orangered}{11}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 11 } = \color{blue}{ 77 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&1&4&-7&1\\& & 7& \color{blue}{77} & \\ \hline &1&\color{blue}{11}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 77 } = \color{orangered}{ 70 } $
$$ \begin{array}{c|rrrr}7&1&4&\color{orangered}{ -7 }&1\\& & 7& \color{orangered}{77} & \\ \hline &1&11&\color{orangered}{70}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 70 } = \color{blue}{ 490 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&1&4&-7&1\\& & 7& 77& \color{blue}{490} \\ \hline &1&11&\color{blue}{70}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 490 } = \color{orangered}{ 491 } $
$$ \begin{array}{c|rrrr}7&1&4&-7&\color{orangered}{ 1 }\\& & 7& 77& \color{orangered}{490} \\ \hline &\color{blue}{1}&\color{blue}{11}&\color{blue}{70}&\color{orangered}{491} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+11x+70 } $ with a remainder of $ \color{red}{ 491 } $.