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