Divide using synthetic division $$ ( x^{3}+5x^{2}+14x+23 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&1&5&14&23\\& & -2& -6& \color{black}{-16} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{8}&\color{orangered}{7} \end{array} $$The solution is:
$$ \frac{ x^{3}+5x^{2}+14x+23 }{ x+2 } = \color{blue}{x^{2}+3x+8} ~+~ \frac{ \color{red}{ 7 } }{ x+2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&5&14&23\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 1 }&5&14&23\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 1 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&5&14&23\\& & \color{blue}{-2} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}-2&1&\color{orangered}{ 5 }&14&23\\& & \color{orangered}{-2} & & \\ \hline &1&\color{orangered}{3}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 3 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&5&14&23\\& & -2& \color{blue}{-6} & \\ \hline &1&\color{blue}{3}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 14 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrr}-2&1&5&\color{orangered}{ 14 }&23\\& & -2& \color{orangered}{-6} & \\ \hline &1&3&\color{orangered}{8}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 8 } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&5&14&23\\& & -2& -6& \color{blue}{-16} \\ \hline &1&3&\color{blue}{8}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 23 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrr}-2&1&5&14&\color{orangered}{ 23 }\\& & -2& -6& \color{orangered}{-16} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{8}&\color{orangered}{7} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+3x+8 } $ with a remainder of $ \color{red}{ 7 } $.