Divide using synthetic division $$ ( x^{4}-2401 ) \div ( x-7 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}7&1&0&0&0&-2401\\& & 7& 49& 343& \color{black}{2401} \\ \hline &\color{blue}{1}&\color{blue}{7}&\color{blue}{49}&\color{blue}{343}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{4}-2401 }{ x-7 } = \color{blue}{x^{3}+7x^{2}+49x+343} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&1&0&0&0&-2401\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}7&\color{orangered}{ 1 }&0&0&0&-2401\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&1&0&0&0&-2401\\& & \color{blue}{7} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 7 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrr}7&1&\color{orangered}{ 0 }&0&0&-2401\\& & \color{orangered}{7} & & & \\ \hline &1&\color{orangered}{7}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&1&0&0&0&-2401\\& & 7& \color{blue}{49} & & \\ \hline &1&\color{blue}{7}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 49 } = \color{orangered}{ 49 } $
$$ \begin{array}{c|rrrrr}7&1&0&\color{orangered}{ 0 }&0&-2401\\& & 7& \color{orangered}{49} & & \\ \hline &1&7&\color{orangered}{49}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 49 } = \color{blue}{ 343 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&1&0&0&0&-2401\\& & 7& 49& \color{blue}{343} & \\ \hline &1&7&\color{blue}{49}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 343 } = \color{orangered}{ 343 } $
$$ \begin{array}{c|rrrrr}7&1&0&0&\color{orangered}{ 0 }&-2401\\& & 7& 49& \color{orangered}{343} & \\ \hline &1&7&49&\color{orangered}{343}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 343 } = \color{blue}{ 2401 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&1&0&0&0&-2401\\& & 7& 49& 343& \color{blue}{2401} \\ \hline &1&7&49&\color{blue}{343}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -2401 } + \color{orangered}{ 2401 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}7&1&0&0&0&\color{orangered}{ -2401 }\\& & 7& 49& 343& \color{orangered}{2401} \\ \hline &\color{blue}{1}&\color{blue}{7}&\color{blue}{49}&\color{blue}{343}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+7x^{2}+49x+343 } $ with a remainder of $ \color{red}{ 0 } $.