Divide using synthetic division $$ ( x^{4}+7x^{3}-21x-7 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&1&7&0&-21&-7\\& & -2& -10& 20& \color{black}{2} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{-10}&\color{blue}{-1}&\color{orangered}{-5} \end{array} $$The solution is:
$$ \frac{ x^{4}+7x^{3}-21x-7 }{ x+2 } = \color{blue}{x^{3}+5x^{2}-10x-1} \color{red}{~-~} \frac{ \color{red}{ 5 } }{ x+2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&7&0&-21&-7\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 1 }&7&0&-21&-7\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 1 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&7&0&-21&-7\\& & \color{blue}{-2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}-2&1&\color{orangered}{ 7 }&0&-21&-7\\& & \color{orangered}{-2} & & & \\ \hline &1&\color{orangered}{5}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 5 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&7&0&-21&-7\\& & -2& \color{blue}{-10} & & \\ \hline &1&\color{blue}{5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrr}-2&1&7&\color{orangered}{ 0 }&-21&-7\\& & -2& \color{orangered}{-10} & & \\ \hline &1&5&\color{orangered}{-10}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&7&0&-21&-7\\& & -2& -10& \color{blue}{20} & \\ \hline &1&5&\color{blue}{-10}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -21 } + \color{orangered}{ 20 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}-2&1&7&0&\color{orangered}{ -21 }&-7\\& & -2& -10& \color{orangered}{20} & \\ \hline &1&5&-10&\color{orangered}{-1}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&7&0&-21&-7\\& & -2& -10& 20& \color{blue}{2} \\ \hline &1&5&-10&\color{blue}{-1}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 2 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}-2&1&7&0&-21&\color{orangered}{ -7 }\\& & -2& -10& 20& \color{orangered}{2} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{-10}&\color{blue}{-1}&\color{orangered}{-5} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+5x^{2}-10x-1 } $ with a remainder of $ \color{red}{ -5 } $.