Divide using synthetic division $$ ( 2x^{3}+9x+7 ) \div ( x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-1&2&0&9&7\\& & -2& 2& \color{black}{-11} \\ \hline &\color{blue}{2}&\color{blue}{-2}&\color{blue}{11}&\color{orangered}{-4} \end{array} $$The solution is:
$$ \frac{ 2x^{3}+9x+7 }{ x+1 } = \color{blue}{2x^{2}-2x+11} \color{red}{~-~} \frac{ \color{red}{ 4 } }{ x+1 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 1 = 0 $ ( $ x = \color{blue}{ -1 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&2&0&9&7\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-1&\color{orangered}{ 2 }&0&9&7\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 2 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&2&0&9&7\\& & \color{blue}{-2} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}-1&2&\color{orangered}{ 0 }&9&7\\& & \color{orangered}{-2} & & \\ \hline &2&\color{orangered}{-2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&2&0&9&7\\& & -2& \color{blue}{2} & \\ \hline &2&\color{blue}{-2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ 2 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrr}-1&2&0&\color{orangered}{ 9 }&7\\& & -2& \color{orangered}{2} & \\ \hline &2&-2&\color{orangered}{11}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 11 } = \color{blue}{ -11 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&2&0&9&7\\& & -2& 2& \color{blue}{-11} \\ \hline &2&-2&\color{blue}{11}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -11 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}-1&2&0&9&\color{orangered}{ 7 }\\& & -2& 2& \color{orangered}{-11} \\ \hline &\color{blue}{2}&\color{blue}{-2}&\color{blue}{11}&\color{orangered}{-4} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-2x+11 } $ with a remainder of $ \color{red}{ -4 } $.