Divide using synthetic division $$ ( x^{4}-31x^{2}+29x+5 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}5&1&0&-31&29&5\\& & 5& 25& -30& \color{black}{-5} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{-6}&\color{blue}{-1}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{4}-31x^{2}+29x+5 }{ x-5 } = \color{blue}{x^{3}+5x^{2}-6x-1} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&0&-31&29&5\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}5&\color{orangered}{ 1 }&0&-31&29&5\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&0&-31&29&5\\& & \color{blue}{5} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 5 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}5&1&\color{orangered}{ 0 }&-31&29&5\\& & \color{orangered}{5} & & & \\ \hline &1&\color{orangered}{5}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 5 } = \color{blue}{ 25 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&0&-31&29&5\\& & 5& \color{blue}{25} & & \\ \hline &1&\color{blue}{5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -31 } + \color{orangered}{ 25 } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}5&1&0&\color{orangered}{ -31 }&29&5\\& & 5& \color{orangered}{25} & & \\ \hline &1&5&\color{orangered}{-6}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ -30 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&0&-31&29&5\\& & 5& 25& \color{blue}{-30} & \\ \hline &1&5&\color{blue}{-6}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 29 } + \color{orangered}{ \left( -30 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}5&1&0&-31&\color{orangered}{ 29 }&5\\& & 5& 25& \color{orangered}{-30} & \\ \hline &1&5&-6&\color{orangered}{-1}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&0&-31&29&5\\& & 5& 25& -30& \color{blue}{-5} \\ \hline &1&5&-6&\color{blue}{-1}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}5&1&0&-31&29&\color{orangered}{ 5 }\\& & 5& 25& -30& \color{orangered}{-5} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{-6}&\color{blue}{-1}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+5x^{2}-6x-1 } $ with a remainder of $ \color{red}{ 0 } $.