Divide using synthetic division $$ ( 8x^{4}+22x^{3}-41x^{2}-x+12 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-4&8&22&-41&-1&12\\& & -32& 40& 4& \color{black}{-12} \\ \hline &\color{blue}{8}&\color{blue}{-10}&\color{blue}{-1}&\color{blue}{3}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 8x^{4}+22x^{3}-41x^{2}-x+12 }{ x+4 } = \color{blue}{8x^{3}-10x^{2}-x+3} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 4 = 0 $ ( $ x = \color{blue}{ -4 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&8&22&-41&-1&12\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-4&\color{orangered}{ 8 }&22&-41&-1&12\\& & & & & \\ \hline &\color{orangered}{8}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 8 } = \color{blue}{ -32 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&8&22&-41&-1&12\\& & \color{blue}{-32} & & & \\ \hline &\color{blue}{8}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 22 } + \color{orangered}{ \left( -32 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrr}-4&8&\color{orangered}{ 22 }&-41&-1&12\\& & \color{orangered}{-32} & & & \\ \hline &8&\color{orangered}{-10}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 40 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&8&22&-41&-1&12\\& & -32& \color{blue}{40} & & \\ \hline &8&\color{blue}{-10}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -41 } + \color{orangered}{ 40 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}-4&8&22&\color{orangered}{ -41 }&-1&12\\& & -32& \color{orangered}{40} & & \\ \hline &8&-10&\color{orangered}{-1}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&8&22&-41&-1&12\\& & -32& 40& \color{blue}{4} & \\ \hline &8&-10&\color{blue}{-1}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 4 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}-4&8&22&-41&\color{orangered}{ -1 }&12\\& & -32& 40& \color{orangered}{4} & \\ \hline &8&-10&-1&\color{orangered}{3}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 3 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&8&22&-41&-1&12\\& & -32& 40& 4& \color{blue}{-12} \\ \hline &8&-10&-1&\color{blue}{3}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-4&8&22&-41&-1&\color{orangered}{ 12 }\\& & -32& 40& 4& \color{orangered}{-12} \\ \hline &\color{blue}{8}&\color{blue}{-10}&\color{blue}{-1}&\color{blue}{3}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 8x^{3}-10x^{2}-x+3 } $ with a remainder of $ \color{red}{ 0 } $.