Divide using synthetic division $$ ( 8x^{4}-64x^{3}+x-28 ) \div ( x-8 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}8&8&-64&0&1&-28\\& & 64& 0& 0& \color{black}{8} \\ \hline &\color{blue}{8}&\color{blue}{0}&\color{blue}{0}&\color{blue}{1}&\color{orangered}{-20} \end{array} $$The solution is:
$$ \frac{ 8x^{4}-64x^{3}+x-28 }{ x-8 } = \color{blue}{8x^{3}+1} \color{red}{~-~} \frac{ \color{red}{ 20 } }{ x-8 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{8}&8&-64&0&1&-28\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}8&\color{orangered}{ 8 }&-64&0&1&-28\\& & & & & \\ \hline &\color{orangered}{8}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 8 } = \color{blue}{ 64 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{8}&8&-64&0&1&-28\\& & \color{blue}{64} & & & \\ \hline &\color{blue}{8}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -64 } + \color{orangered}{ 64 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}8&8&\color{orangered}{ -64 }&0&1&-28\\& & \color{orangered}{64} & & & \\ \hline &8&\color{orangered}{0}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{8}&8&-64&0&1&-28\\& & 64& \color{blue}{0} & & \\ \hline &8&\color{blue}{0}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}8&8&-64&\color{orangered}{ 0 }&1&-28\\& & 64& \color{orangered}{0} & & \\ \hline &8&0&\color{orangered}{0}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{8}&8&-64&0&1&-28\\& & 64& 0& \color{blue}{0} & \\ \hline &8&0&\color{blue}{0}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 0 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}8&8&-64&0&\color{orangered}{ 1 }&-28\\& & 64& 0& \color{orangered}{0} & \\ \hline &8&0&0&\color{orangered}{1}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 1 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{8}&8&-64&0&1&-28\\& & 64& 0& 0& \color{blue}{8} \\ \hline &8&0&0&\color{blue}{1}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -28 } + \color{orangered}{ 8 } = \color{orangered}{ -20 } $
$$ \begin{array}{c|rrrrr}8&8&-64&0&1&\color{orangered}{ -28 }\\& & 64& 0& 0& \color{orangered}{8} \\ \hline &\color{blue}{8}&\color{blue}{0}&\color{blue}{0}&\color{blue}{1}&\color{orangered}{-20} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 8x^{3}+1 } $ with a remainder of $ \color{red}{ -20 } $.