Divide using synthetic division $$ ( x^{4}+10x^{3}+12x^{2}-24x+40 ) \div ( x+8 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-8&1&10&12&-24&40\\& & -8& -16& 32& \color{black}{-64} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{-4}&\color{blue}{8}&\color{orangered}{-24} \end{array} $$The solution is:
$$ \frac{ x^{4}+10x^{3}+12x^{2}-24x+40 }{ x+8 } = \color{blue}{x^{3}+2x^{2}-4x+8} \color{red}{~-~} \frac{ \color{red}{ 24 } }{ x+8 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 8 = 0 $ ( $ x = \color{blue}{ -8 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-8}&1&10&12&-24&40\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-8&\color{orangered}{ 1 }&10&12&-24&40\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ 1 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-8}&1&10&12&-24&40\\& & \color{blue}{-8} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}-8&1&\color{orangered}{ 10 }&12&-24&40\\& & \color{orangered}{-8} & & & \\ \hline &1&\color{orangered}{2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ 2 } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-8}&1&10&12&-24&40\\& & -8& \color{blue}{-16} & & \\ \hline &1&\color{blue}{2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}-8&1&10&\color{orangered}{ 12 }&-24&40\\& & -8& \color{orangered}{-16} & & \\ \hline &1&2&\color{orangered}{-4}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 32 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-8}&1&10&12&-24&40\\& & -8& -16& \color{blue}{32} & \\ \hline &1&2&\color{blue}{-4}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -24 } + \color{orangered}{ 32 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}-8&1&10&12&\color{orangered}{ -24 }&40\\& & -8& -16& \color{orangered}{32} & \\ \hline &1&2&-4&\color{orangered}{8}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ 8 } = \color{blue}{ -64 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-8}&1&10&12&-24&40\\& & -8& -16& 32& \color{blue}{-64} \\ \hline &1&2&-4&\color{blue}{8}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 40 } + \color{orangered}{ \left( -64 \right) } = \color{orangered}{ -24 } $
$$ \begin{array}{c|rrrrr}-8&1&10&12&-24&\color{orangered}{ 40 }\\& & -8& -16& 32& \color{orangered}{-64} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{-4}&\color{blue}{8}&\color{orangered}{-24} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+2x^{2}-4x+8 } $ with a remainder of $ \color{red}{ -24 } $.