Divide using synthetic division $$ ( x^{4}+11x^{2}+29x-7 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-5&1&0&11&29&-7\\& & -5& 25& -180& \color{black}{755} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{36}&\color{blue}{-151}&\color{orangered}{748} \end{array} $$The solution is:
$$ \frac{ x^{4}+11x^{2}+29x-7 }{ x+5 } = \color{blue}{x^{3}-5x^{2}+36x-151} ~+~ \frac{ \color{red}{ 748 } }{ x+5 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&1&0&11&29&-7\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-5&\color{orangered}{ 1 }&0&11&29&-7\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&1&0&11&29&-7\\& & \color{blue}{-5} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}-5&1&\color{orangered}{ 0 }&11&29&-7\\& & \color{orangered}{-5} & & & \\ \hline &1&\color{orangered}{-5}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 25 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&1&0&11&29&-7\\& & -5& \color{blue}{25} & & \\ \hline &1&\color{blue}{-5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ 25 } = \color{orangered}{ 36 } $
$$ \begin{array}{c|rrrrr}-5&1&0&\color{orangered}{ 11 }&29&-7\\& & -5& \color{orangered}{25} & & \\ \hline &1&-5&\color{orangered}{36}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 36 } = \color{blue}{ -180 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&1&0&11&29&-7\\& & -5& 25& \color{blue}{-180} & \\ \hline &1&-5&\color{blue}{36}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 29 } + \color{orangered}{ \left( -180 \right) } = \color{orangered}{ -151 } $
$$ \begin{array}{c|rrrrr}-5&1&0&11&\color{orangered}{ 29 }&-7\\& & -5& 25& \color{orangered}{-180} & \\ \hline &1&-5&36&\color{orangered}{-151}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -151 \right) } = \color{blue}{ 755 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&1&0&11&29&-7\\& & -5& 25& -180& \color{blue}{755} \\ \hline &1&-5&36&\color{blue}{-151}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 755 } = \color{orangered}{ 748 } $
$$ \begin{array}{c|rrrrr}-5&1&0&11&29&\color{orangered}{ -7 }\\& & -5& 25& -180& \color{orangered}{755} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{36}&\color{blue}{-151}&\color{orangered}{748} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-5x^{2}+36x-151 } $ with a remainder of $ \color{red}{ 748 } $.