Divide using synthetic division $$ ( x^{4}+4x^{3}-2x^{2}-x+22 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&1&4&-2&-1&22\\& & -2& -4& 12& \color{black}{-22} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{-6}&\color{blue}{11}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{4}+4x^{3}-2x^{2}-x+22 }{ x+2 } = \color{blue}{x^{3}+2x^{2}-6x+11} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&4&-2&-1&22\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 1 }&4&-2&-1&22\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&4&-2&-1&22\\& & \color{blue}{-2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}-2&1&\color{orangered}{ 4 }&-2&-1&22\\& & \color{orangered}{-2} & & & \\ \hline &1&\color{orangered}{2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 2 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&4&-2&-1&22\\& & -2& \color{blue}{-4} & & \\ \hline &1&\color{blue}{2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}-2&1&4&\color{orangered}{ -2 }&-1&22\\& & -2& \color{orangered}{-4} & & \\ \hline &1&2&\color{orangered}{-6}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&4&-2&-1&22\\& & -2& -4& \color{blue}{12} & \\ \hline &1&2&\color{blue}{-6}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 12 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrrr}-2&1&4&-2&\color{orangered}{ -1 }&22\\& & -2& -4& \color{orangered}{12} & \\ \hline &1&2&-6&\color{orangered}{11}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 11 } = \color{blue}{ -22 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&4&-2&-1&22\\& & -2& -4& 12& \color{blue}{-22} \\ \hline &1&2&-6&\color{blue}{11}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 22 } + \color{orangered}{ \left( -22 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-2&1&4&-2&-1&\color{orangered}{ 22 }\\& & -2& -4& 12& \color{orangered}{-22} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{-6}&\color{blue}{11}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+2x^{2}-6x+11 } $ with a remainder of $ \color{red}{ 0 } $.