Divide using synthetic division $$ ( 7x^{3}+2x^{2}-16 ) \div ( x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-1&7&2&0&-16\\& & -7& 5& \color{black}{-5} \\ \hline &\color{blue}{7}&\color{blue}{-5}&\color{blue}{5}&\color{orangered}{-21} \end{array} $$The solution is:
$$ \frac{ 7x^{3}+2x^{2}-16 }{ x+1 } = \color{blue}{7x^{2}-5x+5} \color{red}{~-~} \frac{ \color{red}{ 21 } }{ x+1 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&7&2&0&-16\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-1&\color{orangered}{ 7 }&2&0&-16\\& & & & \\ \hline &\color{orangered}{7}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&7&2&0&-16\\& & \color{blue}{-7} & & \\ \hline &\color{blue}{7}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -7 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}-1&7&\color{orangered}{ 2 }&0&-16\\& & \color{orangered}{-7} & & \\ \hline &7&\color{orangered}{-5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&7&2&0&-16\\& & -7& \color{blue}{5} & \\ \hline &7&\color{blue}{-5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 5 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}-1&7&2&\color{orangered}{ 0 }&-16\\& & -7& \color{orangered}{5} & \\ \hline &7&-5&\color{orangered}{5}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 5 } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&7&2&0&-16\\& & -7& 5& \color{blue}{-5} \\ \hline &7&-5&\color{blue}{5}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -21 } $
$$ \begin{array}{c|rrrr}-1&7&2&0&\color{orangered}{ -16 }\\& & -7& 5& \color{orangered}{-5} \\ \hline &\color{blue}{7}&\color{blue}{-5}&\color{blue}{5}&\color{orangered}{-21} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 7x^{2}-5x+5 } $ with a remainder of $ \color{red}{ -21 } $.