Divide using synthetic division $$ ( x^{4}-6x^{3}-8x+48 ) \div ( x-6 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}6&1&-6&0&-8&48\\& & 6& 0& 0& \color{black}{-48} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{0}&\color{blue}{-8}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{4}-6x^{3}-8x+48 }{ x-6 } = \color{blue}{x^{3}-8} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -6 = 0 $ ( $ x = \color{blue}{ 6 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&1&-6&0&-8&48\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}6&\color{orangered}{ 1 }&-6&0&-8&48\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 1 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&1&-6&0&-8&48\\& & \color{blue}{6} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 6 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}6&1&\color{orangered}{ -6 }&0&-8&48\\& & \color{orangered}{6} & & & \\ \hline &1&\color{orangered}{0}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&1&-6&0&-8&48\\& & 6& \color{blue}{0} & & \\ \hline &1&\color{blue}{0}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}6&1&-6&\color{orangered}{ 0 }&-8&48\\& & 6& \color{orangered}{0} & & \\ \hline &1&0&\color{orangered}{0}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&1&-6&0&-8&48\\& & 6& 0& \color{blue}{0} & \\ \hline &1&0&\color{blue}{0}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 0 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}6&1&-6&0&\color{orangered}{ -8 }&48\\& & 6& 0& \color{orangered}{0} & \\ \hline &1&0&0&\color{orangered}{-8}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -48 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&1&-6&0&-8&48\\& & 6& 0& 0& \color{blue}{-48} \\ \hline &1&0&0&\color{blue}{-8}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 48 } + \color{orangered}{ \left( -48 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}6&1&-6&0&-8&\color{orangered}{ 48 }\\& & 6& 0& 0& \color{orangered}{-48} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{0}&\color{blue}{-8}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-8 } $ with a remainder of $ \color{red}{ 0 } $.