Divide using synthetic division $$ ( x^{4}-2x^{3}-29x^{2}-43x+8 ) \div ( x-7 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}7&1&-2&-29&-43&8\\& & 7& 35& 42& \color{black}{-7} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{6}&\color{blue}{-1}&\color{orangered}{1} \end{array} $$The solution is:
$$ \frac{ x^{4}-2x^{3}-29x^{2}-43x+8 }{ x-7 } = \color{blue}{x^{3}+5x^{2}+6x-1} ~+~ \frac{ \color{red}{ 1 } }{ x-7 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&1&-2&-29&-43&8\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}7&\color{orangered}{ 1 }&-2&-29&-43&8\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&1&-2&-29&-43&8\\& & \color{blue}{7} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 7 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}7&1&\color{orangered}{ -2 }&-29&-43&8\\& & \color{orangered}{7} & & & \\ \hline &1&\color{orangered}{5}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 5 } = \color{blue}{ 35 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&1&-2&-29&-43&8\\& & 7& \color{blue}{35} & & \\ \hline &1&\color{blue}{5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -29 } + \color{orangered}{ 35 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}7&1&-2&\color{orangered}{ -29 }&-43&8\\& & 7& \color{orangered}{35} & & \\ \hline &1&5&\color{orangered}{6}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 6 } = \color{blue}{ 42 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&1&-2&-29&-43&8\\& & 7& 35& \color{blue}{42} & \\ \hline &1&5&\color{blue}{6}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -43 } + \color{orangered}{ 42 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}7&1&-2&-29&\color{orangered}{ -43 }&8\\& & 7& 35& \color{orangered}{42} & \\ \hline &1&5&6&\color{orangered}{-1}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -7 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&1&-2&-29&-43&8\\& & 7& 35& 42& \color{blue}{-7} \\ \hline &1&5&6&\color{blue}{-1}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -7 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}7&1&-2&-29&-43&\color{orangered}{ 8 }\\& & 7& 35& 42& \color{orangered}{-7} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{6}&\color{blue}{-1}&\color{orangered}{1} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+5x^{2}+6x-1 } $ with a remainder of $ \color{red}{ 1 } $.