Divide using synthetic division $$ ( x^{3}+15x^{2}+47x-38 ) \div ( x+6 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-6&1&15&47&-38\\& & -6& -54& \color{black}{42} \\ \hline &\color{blue}{1}&\color{blue}{9}&\color{blue}{-7}&\color{orangered}{4} \end{array} $$The solution is:
$$ \frac{ x^{3}+15x^{2}+47x-38 }{ x+6 } = \color{blue}{x^{2}+9x-7} ~+~ \frac{ \color{red}{ 4 } }{ x+6 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 6 = 0 $ ( $ x = \color{blue}{ -6 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-6}&1&15&47&-38\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-6&\color{orangered}{ 1 }&15&47&-38\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -6 } \cdot \color{blue}{ 1 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-6}&1&15&47&-38\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrr}-6&1&\color{orangered}{ 15 }&47&-38\\& & \color{orangered}{-6} & & \\ \hline &1&\color{orangered}{9}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -6 } \cdot \color{blue}{ 9 } = \color{blue}{ -54 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-6}&1&15&47&-38\\& & -6& \color{blue}{-54} & \\ \hline &1&\color{blue}{9}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 47 } + \color{orangered}{ \left( -54 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrr}-6&1&15&\color{orangered}{ 47 }&-38\\& & -6& \color{orangered}{-54} & \\ \hline &1&9&\color{orangered}{-7}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -6 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 42 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-6}&1&15&47&-38\\& & -6& -54& \color{blue}{42} \\ \hline &1&9&\color{blue}{-7}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -38 } + \color{orangered}{ 42 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}-6&1&15&47&\color{orangered}{ -38 }\\& & -6& -54& \color{orangered}{42} \\ \hline &\color{blue}{1}&\color{blue}{9}&\color{blue}{-7}&\color{orangered}{4} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+9x-7 } $ with a remainder of $ \color{red}{ 4 } $.