Divide using synthetic division $$ ( x^{4}-5x^{3}-22x^{2}+5x+6 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&1&-5&-22&5&6\\& & -3& 24& -6& \color{black}{3} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{blue}{2}&\color{blue}{-1}&\color{orangered}{9} \end{array} $$The solution is:
$$ \frac{ x^{4}-5x^{3}-22x^{2}+5x+6 }{ x+3 } = \color{blue}{x^{3}-8x^{2}+2x-1} ~+~ \frac{ \color{red}{ 9 } }{ x+3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 3 = 0 $ ( $ x = \color{blue}{ -3 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-5&-22&5&6\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 1 }&-5&-22&5&6\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 1 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-5&-22&5&6\\& & \color{blue}{-3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}-3&1&\color{orangered}{ -5 }&-22&5&6\\& & \color{orangered}{-3} & & & \\ \hline &1&\color{orangered}{-8}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-5&-22&5&6\\& & -3& \color{blue}{24} & & \\ \hline &1&\color{blue}{-8}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -22 } + \color{orangered}{ 24 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}-3&1&-5&\color{orangered}{ -22 }&5&6\\& & -3& \color{orangered}{24} & & \\ \hline &1&-8&\color{orangered}{2}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 2 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-5&-22&5&6\\& & -3& 24& \color{blue}{-6} & \\ \hline &1&-8&\color{blue}{2}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}-3&1&-5&-22&\color{orangered}{ 5 }&6\\& & -3& 24& \color{orangered}{-6} & \\ \hline &1&-8&2&\color{orangered}{-1}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-5&-22&5&6\\& & -3& 24& -6& \color{blue}{3} \\ \hline &1&-8&2&\color{blue}{-1}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 3 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrr}-3&1&-5&-22&5&\color{orangered}{ 6 }\\& & -3& 24& -6& \color{orangered}{3} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{blue}{2}&\color{blue}{-1}&\color{orangered}{9} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-8x^{2}+2x-1 } $ with a remainder of $ \color{red}{ 9 } $.