Divide using synthetic division $$ ( x^{3}+12x^{2}+25x+50 ) \div ( x+10 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-10&1&12&25&50\\& & -10& -20& \color{black}{-50} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{5}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{3}+12x^{2}+25x+50 }{ x+10 } = \color{blue}{x^{2}+2x+5} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 10 = 0 $ ( $ x = \color{blue}{ -10 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&1&12&25&50\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-10&\color{orangered}{ 1 }&12&25&50\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ 1 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&1&12&25&50\\& & \color{blue}{-10} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}-10&1&\color{orangered}{ 12 }&25&50\\& & \color{orangered}{-10} & & \\ \hline &1&\color{orangered}{2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ 2 } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&1&12&25&50\\& & -10& \color{blue}{-20} & \\ \hline &1&\color{blue}{2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 25 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}-10&1&12&\color{orangered}{ 25 }&50\\& & -10& \color{orangered}{-20} & \\ \hline &1&2&\color{orangered}{5}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ 5 } = \color{blue}{ -50 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&1&12&25&50\\& & -10& -20& \color{blue}{-50} \\ \hline &1&2&\color{blue}{5}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 50 } + \color{orangered}{ \left( -50 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}-10&1&12&25&\color{orangered}{ 50 }\\& & -10& -20& \color{orangered}{-50} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{5}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+2x+5 } $ with a remainder of $ \color{red}{ 0 } $.