Divide using synthetic division $$ ( x^{3}-46x-14 ) \div ( x-7 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}7&1&0&-46&-14\\& & 7& 49& \color{black}{21} \\ \hline &\color{blue}{1}&\color{blue}{7}&\color{blue}{3}&\color{orangered}{7} \end{array} $$The solution is:
$$ \frac{ x^{3}-46x-14 }{ x-7 } = \color{blue}{x^{2}+7x+3} ~+~ \frac{ \color{red}{ 7 } }{ x-7 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{7}&1&0&-46&-14\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}7&\color{orangered}{ 1 }&0&-46&-14\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&1&0&-46&-14\\& & \color{blue}{7} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 7 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrr}7&1&\color{orangered}{ 0 }&-46&-14\\& & \color{orangered}{7} & & \\ \hline &1&\color{orangered}{7}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 7 } = \color{blue}{ 49 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&1&0&-46&-14\\& & 7& \color{blue}{49} & \\ \hline &1&\color{blue}{7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -46 } + \color{orangered}{ 49 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}7&1&0&\color{orangered}{ -46 }&-14\\& & 7& \color{orangered}{49} & \\ \hline &1&7&\color{orangered}{3}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 3 } = \color{blue}{ 21 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&1&0&-46&-14\\& & 7& 49& \color{blue}{21} \\ \hline &1&7&\color{blue}{3}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 21 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrr}7&1&0&-46&\color{orangered}{ -14 }\\& & 7& 49& \color{orangered}{21} \\ \hline &\color{blue}{1}&\color{blue}{7}&\color{blue}{3}&\color{orangered}{7} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+7x+3 } $ with a remainder of $ \color{red}{ 7 } $.