Divide using synthetic division $$ ( x^{3}-62x-20 ) \div ( x-8 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}8&1&0&-62&-20\\& & 8& 64& \color{black}{16} \\ \hline &\color{blue}{1}&\color{blue}{8}&\color{blue}{2}&\color{orangered}{-4} \end{array} $$The solution is:
$$ \frac{ x^{3}-62x-20 }{ x-8 } = \color{blue}{x^{2}+8x+2} \color{red}{~-~} \frac{ \color{red}{ 4 } }{ x-8 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{8}&1&0&-62&-20\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}8&\color{orangered}{ 1 }&0&-62&-20\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{8}&1&0&-62&-20\\& & \color{blue}{8} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 8 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrr}8&1&\color{orangered}{ 0 }&-62&-20\\& & \color{orangered}{8} & & \\ \hline &1&\color{orangered}{8}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 8 } = \color{blue}{ 64 } $.
$$ \begin{array}{c|rrrr}\color{blue}{8}&1&0&-62&-20\\& & 8& \color{blue}{64} & \\ \hline &1&\color{blue}{8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -62 } + \color{orangered}{ 64 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}8&1&0&\color{orangered}{ -62 }&-20\\& & 8& \color{orangered}{64} & \\ \hline &1&8&\color{orangered}{2}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 2 } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{8}&1&0&-62&-20\\& & 8& 64& \color{blue}{16} \\ \hline &1&8&\color{blue}{2}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 16 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}8&1&0&-62&\color{orangered}{ -20 }\\& & 8& 64& \color{orangered}{16} \\ \hline &\color{blue}{1}&\color{blue}{8}&\color{blue}{2}&\color{orangered}{-4} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+8x+2 } $ with a remainder of $ \color{red}{ -4 } $.