Divide using synthetic division $$ ( x^{4}+3x^{3}+3x^{2}-6x-7 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}5&1&3&3&-6&-7\\& & 5& 40& 215& \color{black}{1045} \\ \hline &\color{blue}{1}&\color{blue}{8}&\color{blue}{43}&\color{blue}{209}&\color{orangered}{1038} \end{array} $$The solution is:
$$ \frac{ x^{4}+3x^{3}+3x^{2}-6x-7 }{ x-5 } = \color{blue}{x^{3}+8x^{2}+43x+209} ~+~ \frac{ \color{red}{ 1038 } }{ x-5 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&3&3&-6&-7\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}5&\color{orangered}{ 1 }&3&3&-6&-7\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 1 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&3&3&-6&-7\\& & \color{blue}{5} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 5 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}5&1&\color{orangered}{ 3 }&3&-6&-7\\& & \color{orangered}{5} & & & \\ \hline &1&\color{orangered}{8}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 8 } = \color{blue}{ 40 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&3&3&-6&-7\\& & 5& \color{blue}{40} & & \\ \hline &1&\color{blue}{8}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 40 } = \color{orangered}{ 43 } $
$$ \begin{array}{c|rrrrr}5&1&3&\color{orangered}{ 3 }&-6&-7\\& & 5& \color{orangered}{40} & & \\ \hline &1&8&\color{orangered}{43}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 43 } = \color{blue}{ 215 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&3&3&-6&-7\\& & 5& 40& \color{blue}{215} & \\ \hline &1&8&\color{blue}{43}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 215 } = \color{orangered}{ 209 } $
$$ \begin{array}{c|rrrrr}5&1&3&3&\color{orangered}{ -6 }&-7\\& & 5& 40& \color{orangered}{215} & \\ \hline &1&8&43&\color{orangered}{209}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 209 } = \color{blue}{ 1045 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&3&3&-6&-7\\& & 5& 40& 215& \color{blue}{1045} \\ \hline &1&8&43&\color{blue}{209}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 1045 } = \color{orangered}{ 1038 } $
$$ \begin{array}{c|rrrrr}5&1&3&3&-6&\color{orangered}{ -7 }\\& & 5& 40& 215& \color{orangered}{1045} \\ \hline &\color{blue}{1}&\color{blue}{8}&\color{blue}{43}&\color{blue}{209}&\color{orangered}{1038} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+8x^{2}+43x+209 } $ with a remainder of $ \color{red}{ 1038 } $.