Divide using synthetic division $$ ( x^{3}+x^{2}+25 ) \div ( x+6 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-6&1&1&0&25\\& & -6& 30& \color{black}{-180} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{30}&\color{orangered}{-155} \end{array} $$The solution is:
$$ \frac{ x^{3}+x^{2}+25 }{ x+6 } = \color{blue}{x^{2}-5x+30} \color{red}{~-~} \frac{ \color{red}{ 155 } }{ x+6 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 6 = 0 $ ( $ x = \color{blue}{ -6 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-6}&1&1&0&25\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-6&\color{orangered}{ 1 }&1&0&25\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -6 } \cdot \color{blue}{ 1 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-6}&1&1&0&25\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}-6&1&\color{orangered}{ 1 }&0&25\\& & \color{orangered}{-6} & & \\ \hline &1&\color{orangered}{-5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -6 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-6}&1&1&0&25\\& & -6& \color{blue}{30} & \\ \hline &1&\color{blue}{-5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 30 } = \color{orangered}{ 30 } $
$$ \begin{array}{c|rrrr}-6&1&1&\color{orangered}{ 0 }&25\\& & -6& \color{orangered}{30} & \\ \hline &1&-5&\color{orangered}{30}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -6 } \cdot \color{blue}{ 30 } = \color{blue}{ -180 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-6}&1&1&0&25\\& & -6& 30& \color{blue}{-180} \\ \hline &1&-5&\color{blue}{30}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 25 } + \color{orangered}{ \left( -180 \right) } = \color{orangered}{ -155 } $
$$ \begin{array}{c|rrrr}-6&1&1&0&\color{orangered}{ 25 }\\& & -6& 30& \color{orangered}{-180} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{30}&\color{orangered}{-155} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-5x+30 } $ with a remainder of $ \color{red}{ -155 } $.