Divide using synthetic division $$ ( x^{3}-6x^{2}-46 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}5&1&-6&0&-46\\& & 5& -5& \color{black}{-25} \\ \hline &\color{blue}{1}&\color{blue}{-1}&\color{blue}{-5}&\color{orangered}{-71} \end{array} $$The solution is:
$$ \frac{ x^{3}-6x^{2}-46 }{ x-5 } = \color{blue}{x^{2}-x-5} \color{red}{~-~} \frac{ \color{red}{ 71 } }{ x-5 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&-6&0&-46\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}5&\color{orangered}{ 1 }&-6&0&-46\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 1 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&-6&0&-46\\& & \color{blue}{5} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 5 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}5&1&\color{orangered}{ -6 }&0&-46\\& & \color{orangered}{5} & & \\ \hline &1&\color{orangered}{-1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&-6&0&-46\\& & 5& \color{blue}{-5} & \\ \hline &1&\color{blue}{-1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}5&1&-6&\color{orangered}{ 0 }&-46\\& & 5& \color{orangered}{-5} & \\ \hline &1&-1&\color{orangered}{-5}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -25 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&-6&0&-46\\& & 5& -5& \color{blue}{-25} \\ \hline &1&-1&\color{blue}{-5}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -46 } + \color{orangered}{ \left( -25 \right) } = \color{orangered}{ -71 } $
$$ \begin{array}{c|rrrr}5&1&-6&0&\color{orangered}{ -46 }\\& & 5& -5& \color{orangered}{-25} \\ \hline &\color{blue}{1}&\color{blue}{-1}&\color{blue}{-5}&\color{orangered}{-71} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-x-5 } $ with a remainder of $ \color{red}{ -71 } $.