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