Divide using synthetic division $$ ( 2x^{3}+14x^{2}-20x+7 ) \div ( x+6 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-6&2&14&-20&7\\& & -12& -12& \color{black}{192} \\ \hline &\color{blue}{2}&\color{blue}{2}&\color{blue}{-32}&\color{orangered}{199} \end{array} $$The solution is:
$$ \frac{ 2x^{3}+14x^{2}-20x+7 }{ x+6 } = \color{blue}{2x^{2}+2x-32} ~+~ \frac{ \color{red}{ 199 } }{ x+6 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-6}&2&14&-20&7\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-6&\color{orangered}{ 2 }&14&-20&7\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -6 } \cdot \color{blue}{ 2 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-6}&2&14&-20&7\\& & \color{blue}{-12} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 14 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}-6&2&\color{orangered}{ 14 }&-20&7\\& & \color{orangered}{-12} & & \\ \hline &2&\color{orangered}{2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -6 } \cdot \color{blue}{ 2 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-6}&2&14&-20&7\\& & -12& \color{blue}{-12} & \\ \hline &2&\color{blue}{2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -32 } $
$$ \begin{array}{c|rrrr}-6&2&14&\color{orangered}{ -20 }&7\\& & -12& \color{orangered}{-12} & \\ \hline &2&2&\color{orangered}{-32}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -6 } \cdot \color{blue}{ \left( -32 \right) } = \color{blue}{ 192 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-6}&2&14&-20&7\\& & -12& -12& \color{blue}{192} \\ \hline &2&2&\color{blue}{-32}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 192 } = \color{orangered}{ 199 } $
$$ \begin{array}{c|rrrr}-6&2&14&-20&\color{orangered}{ 7 }\\& & -12& -12& \color{orangered}{192} \\ \hline &\color{blue}{2}&\color{blue}{2}&\color{blue}{-32}&\color{orangered}{199} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}+2x-32 } $ with a remainder of $ \color{red}{ 199 } $.