Divide using synthetic division $$ ( x^{3}+5x^{2}+11x+15 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&1&5&11&15\\& & -3& -6& \color{black}{-15} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{5}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{3}+5x^{2}+11x+15 }{ x+3 } = \color{blue}{x^{2}+2x+5} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&5&11&15\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 1 }&5&11&15\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&5&11&15\\& & \color{blue}{-3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}-3&1&\color{orangered}{ 5 }&11&15\\& & \color{orangered}{-3} & & \\ \hline &1&\color{orangered}{2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 2 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&5&11&15\\& & -3& \color{blue}{-6} & \\ \hline &1&\color{blue}{2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}-3&1&5&\color{orangered}{ 11 }&15\\& & -3& \color{orangered}{-6} & \\ \hline &1&2&\color{orangered}{5}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 5 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&5&11&15\\& & -3& -6& \color{blue}{-15} \\ \hline &1&2&\color{blue}{5}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}-3&1&5&11&\color{orangered}{ 15 }\\& & -3& -6& \color{orangered}{-15} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{5}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+2x+5 } $ with a remainder of $ \color{red}{ 0 } $.