Divide using synthetic division $$ ( x^{4}-5x^{3}-23x^{2}+5x+6 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&1&-5&-23&5&6\\& & -3& 24& -3& \color{black}{-6} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{blue}{1}&\color{blue}{2}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{4}-5x^{3}-23x^{2}+5x+6 }{ x+3 } = \color{blue}{x^{3}-8x^{2}+x+2} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-5&-23&5&6\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 1 }&-5&-23&5&6\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-5&-23&5&6\\& & \color{blue}{-3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}-3&1&\color{orangered}{ -5 }&-23&5&6\\& & \color{orangered}{-3} & & & \\ \hline &1&\color{orangered}{-8}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-5&-23&5&6\\& & -3& \color{blue}{24} & & \\ \hline &1&\color{blue}{-8}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -23 } + \color{orangered}{ 24 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}-3&1&-5&\color{orangered}{ -23 }&5&6\\& & -3& \color{orangered}{24} & & \\ \hline &1&-8&\color{orangered}{1}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 1 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-5&-23&5&6\\& & -3& 24& \color{blue}{-3} & \\ \hline &1&-8&\color{blue}{1}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}-3&1&-5&-23&\color{orangered}{ 5 }&6\\& & -3& 24& \color{orangered}{-3} & \\ \hline &1&-8&1&\color{orangered}{2}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 2 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-5&-23&5&6\\& & -3& 24& -3& \color{blue}{-6} \\ \hline &1&-8&1&\color{blue}{2}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-3&1&-5&-23&5&\color{orangered}{ 6 }\\& & -3& 24& -3& \color{orangered}{-6} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{blue}{1}&\color{blue}{2}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-8x^{2}+x+2 } $ with a remainder of $ \color{red}{ 0 } $.