Divide using synthetic division $$ ( x^{3}-12x^{2}-1 ) \div ( x-7 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}7&1&-12&0&-1\\& & 7& -35& \color{black}{-245} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{-35}&\color{orangered}{-246} \end{array} $$The solution is:
$$ \frac{ x^{3}-12x^{2}-1 }{ x-7 } = \color{blue}{x^{2}-5x-35} \color{red}{~-~} \frac{ \color{red}{ 246 } }{ x-7 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -7 = 0 $ ( $ x = \color{blue}{ 7 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{7}&1&-12&0&-1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}7&\color{orangered}{ 1 }&-12&0&-1\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 1 } = \color{blue}{ 7 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&1&-12&0&-1\\& & \color{blue}{7} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 7 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}7&1&\color{orangered}{ -12 }&0&-1\\& & \color{orangered}{7} & & \\ \hline &1&\color{orangered}{-5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -35 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&1&-12&0&-1\\& & 7& \color{blue}{-35} & \\ \hline &1&\color{blue}{-5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -35 \right) } = \color{orangered}{ -35 } $
$$ \begin{array}{c|rrrr}7&1&-12&\color{orangered}{ 0 }&-1\\& & 7& \color{orangered}{-35} & \\ \hline &1&-5&\color{orangered}{-35}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ \left( -35 \right) } = \color{blue}{ -245 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&1&-12&0&-1\\& & 7& -35& \color{blue}{-245} \\ \hline &1&-5&\color{blue}{-35}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -245 \right) } = \color{orangered}{ -246 } $
$$ \begin{array}{c|rrrr}7&1&-12&0&\color{orangered}{ -1 }\\& & 7& -35& \color{orangered}{-245} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{-35}&\color{orangered}{-246} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-5x-35 } $ with a remainder of $ \color{red}{ -246 } $.