Divide using synthetic division $$ ( x^{4}+9x^{3}+13x^{2}-11x-25 ) \div ( x+7 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-7&1&9&13&-11&-25\\& & -7& -14& 7& \color{black}{28} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{-1}&\color{blue}{-4}&\color{orangered}{3} \end{array} $$The solution is:
$$ \frac{ x^{4}+9x^{3}+13x^{2}-11x-25 }{ x+7 } = \color{blue}{x^{3}+2x^{2}-x-4} ~+~ \frac{ \color{red}{ 3 } }{ x+7 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-7}&1&9&13&-11&-25\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-7&\color{orangered}{ 1 }&9&13&-11&-25\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-7}&1&9&13&-11&-25\\& & \color{blue}{-7} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -7 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}-7&1&\color{orangered}{ 9 }&13&-11&-25\\& & \color{orangered}{-7} & & & \\ \hline &1&\color{orangered}{2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -7 } \cdot \color{blue}{ 2 } = \color{blue}{ -14 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-7}&1&9&13&-11&-25\\& & -7& \color{blue}{-14} & & \\ \hline &1&\color{blue}{2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 13 } + \color{orangered}{ \left( -14 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}-7&1&9&\color{orangered}{ 13 }&-11&-25\\& & -7& \color{orangered}{-14} & & \\ \hline &1&2&\color{orangered}{-1}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -7 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 7 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-7}&1&9&13&-11&-25\\& & -7& -14& \color{blue}{7} & \\ \hline &1&2&\color{blue}{-1}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 7 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}-7&1&9&13&\color{orangered}{ -11 }&-25\\& & -7& -14& \color{orangered}{7} & \\ \hline &1&2&-1&\color{orangered}{-4}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -7 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 28 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-7}&1&9&13&-11&-25\\& & -7& -14& 7& \color{blue}{28} \\ \hline &1&2&-1&\color{blue}{-4}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -25 } + \color{orangered}{ 28 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}-7&1&9&13&-11&\color{orangered}{ -25 }\\& & -7& -14& 7& \color{orangered}{28} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{-1}&\color{blue}{-4}&\color{orangered}{3} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+2x^{2}-x-4 } $ with a remainder of $ \color{red}{ 3 } $.