Divide using synthetic division $$ ( x^{4}+6x^{3}+11x^{2}+29x-13 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-5&1&6&11&29&-13\\& & -5& -5& -30& \color{black}{5} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{6}&\color{blue}{-1}&\color{orangered}{-8} \end{array} $$The solution is:
$$ \frac{ x^{4}+6x^{3}+11x^{2}+29x-13 }{ x+5 } = \color{blue}{x^{3}+x^{2}+6x-1} \color{red}{~-~} \frac{ \color{red}{ 8 } }{ x+5 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&1&6&11&29&-13\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-5&\color{orangered}{ 1 }&6&11&29&-13\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&1&6&11&29&-13\\& & \color{blue}{-5} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}-5&1&\color{orangered}{ 6 }&11&29&-13\\& & \color{orangered}{-5} & & & \\ \hline &1&\color{orangered}{1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 1 } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&1&6&11&29&-13\\& & -5& \color{blue}{-5} & & \\ \hline &1&\color{blue}{1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}-5&1&6&\color{orangered}{ 11 }&29&-13\\& & -5& \color{orangered}{-5} & & \\ \hline &1&1&\color{orangered}{6}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 6 } = \color{blue}{ -30 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&1&6&11&29&-13\\& & -5& -5& \color{blue}{-30} & \\ \hline &1&1&\color{blue}{6}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 29 } + \color{orangered}{ \left( -30 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}-5&1&6&11&\color{orangered}{ 29 }&-13\\& & -5& -5& \color{orangered}{-30} & \\ \hline &1&1&6&\color{orangered}{-1}& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&1&6&11&29&-13\\& & -5& -5& -30& \color{blue}{5} \\ \hline &1&1&6&\color{blue}{-1}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -13 } + \color{orangered}{ 5 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}-5&1&6&11&29&\color{orangered}{ -13 }\\& & -5& -5& -30& \color{orangered}{5} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{6}&\color{blue}{-1}&\color{orangered}{-8} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+x^{2}+6x-1 } $ with a remainder of $ \color{red}{ -8 } $.