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