Divide using synthetic division $$ ( x^{3}-14x-8 ) \div ( x+8 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-8&1&0&-14&-8\\& & -8& 64& \color{black}{-400} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{blue}{50}&\color{orangered}{-408} \end{array} $$The solution is:
$$ \frac{ x^{3}-14x-8 }{ x+8 } = \color{blue}{x^{2}-8x+50} \color{red}{~-~} \frac{ \color{red}{ 408 } }{ x+8 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-8}&1&0&-14&-8\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-8&\color{orangered}{ 1 }&0&-14&-8\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-8}&1&0&-14&-8\\& & \color{blue}{-8} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrr}-8&1&\color{orangered}{ 0 }&-14&-8\\& & \color{orangered}{-8} & & \\ \hline &1&\color{orangered}{-8}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 64 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-8}&1&0&-14&-8\\& & -8& \color{blue}{64} & \\ \hline &1&\color{blue}{-8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 64 } = \color{orangered}{ 50 } $
$$ \begin{array}{c|rrrr}-8&1&0&\color{orangered}{ -14 }&-8\\& & -8& \color{orangered}{64} & \\ \hline &1&-8&\color{orangered}{50}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ 50 } = \color{blue}{ -400 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-8}&1&0&-14&-8\\& & -8& 64& \color{blue}{-400} \\ \hline &1&-8&\color{blue}{50}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ \left( -400 \right) } = \color{orangered}{ -408 } $
$$ \begin{array}{c|rrrr}-8&1&0&-14&\color{orangered}{ -8 }\\& & -8& 64& \color{orangered}{-400} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{blue}{50}&\color{orangered}{-408} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-8x+50 } $ with a remainder of $ \color{red}{ -408 } $.