Divide using synthetic division $$ ( x^{3}+7x^{2}+20x+25 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&1&7&20&25\\& & -3& -12& \color{black}{-24} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{8}&\color{orangered}{1} \end{array} $$The solution is:
$$ \frac{ x^{3}+7x^{2}+20x+25 }{ x+3 } = \color{blue}{x^{2}+4x+8} ~+~ \frac{ \color{red}{ 1 } }{ x+3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 3 = 0 $ ( $ x = \color{blue}{ -3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&7&20&25\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 1 }&7&20&25\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 1 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&7&20&25\\& & \color{blue}{-3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}-3&1&\color{orangered}{ 7 }&20&25\\& & \color{orangered}{-3} & & \\ \hline &1&\color{orangered}{4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 4 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&7&20&25\\& & -3& \color{blue}{-12} & \\ \hline &1&\color{blue}{4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrr}-3&1&7&\color{orangered}{ 20 }&25\\& & -3& \color{orangered}{-12} & \\ \hline &1&4&\color{orangered}{8}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 8 } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&7&20&25\\& & -3& -12& \color{blue}{-24} \\ \hline &1&4&\color{blue}{8}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 25 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrr}-3&1&7&20&\color{orangered}{ 25 }\\& & -3& -12& \color{orangered}{-24} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{8}&\color{orangered}{1} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+4x+8 } $ with a remainder of $ \color{red}{ 1 } $.