Divide using synthetic division $$ ( x^{3}-5x^{2}+4x+6 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-5&1&-5&4&6\\& & -5& 50& \color{black}{-270} \\ \hline &\color{blue}{1}&\color{blue}{-10}&\color{blue}{54}&\color{orangered}{-264} \end{array} $$The solution is:
$$ \frac{ x^{3}-5x^{2}+4x+6 }{ x+5 } = \color{blue}{x^{2}-10x+54} \color{red}{~-~} \frac{ \color{red}{ 264 } }{ x+5 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 5 = 0 $ ( $ x = \color{blue}{ -5 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&-5&4&6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-5&\color{orangered}{ 1 }&-5&4&6\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 1 } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&-5&4&6\\& & \color{blue}{-5} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrr}-5&1&\color{orangered}{ -5 }&4&6\\& & \color{orangered}{-5} & & \\ \hline &1&\color{orangered}{-10}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 50 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&-5&4&6\\& & -5& \color{blue}{50} & \\ \hline &1&\color{blue}{-10}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 50 } = \color{orangered}{ 54 } $
$$ \begin{array}{c|rrrr}-5&1&-5&\color{orangered}{ 4 }&6\\& & -5& \color{orangered}{50} & \\ \hline &1&-10&\color{orangered}{54}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 54 } = \color{blue}{ -270 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&-5&4&6\\& & -5& 50& \color{blue}{-270} \\ \hline &1&-10&\color{blue}{54}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -270 \right) } = \color{orangered}{ -264 } $
$$ \begin{array}{c|rrrr}-5&1&-5&4&\color{orangered}{ 6 }\\& & -5& 50& \color{orangered}{-270} \\ \hline &\color{blue}{1}&\color{blue}{-10}&\color{blue}{54}&\color{orangered}{-264} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-10x+54 } $ with a remainder of $ \color{red}{ -264 } $.