Divide using synthetic division $$ ( x^{5}-4x^{4}-25x+100 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}3&1&-4&0&0&-25&100\\& & 3& -3& -9& -27& \color{black}{-156} \\ \hline &\color{blue}{1}&\color{blue}{-1}&\color{blue}{-3}&\color{blue}{-9}&\color{blue}{-52}&\color{orangered}{-56} \end{array} $$The solution is:
$$ \frac{ x^{5}-4x^{4}-25x+100 }{ x-3 } = \color{blue}{x^{4}-x^{3}-3x^{2}-9x-52} \color{red}{~-~} \frac{ \color{red}{ 56 } }{ x-3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-4&0&0&-25&100\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}3&\color{orangered}{ 1 }&-4&0&0&-25&100\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-4&0&0&-25&100\\& & \color{blue}{3} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 3 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrrr}3&1&\color{orangered}{ -4 }&0&0&-25&100\\& & \color{orangered}{3} & & & & \\ \hline &1&\color{orangered}{-1}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-4&0&0&-25&100\\& & 3& \color{blue}{-3} & & & \\ \hline &1&\color{blue}{-1}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrrr}3&1&-4&\color{orangered}{ 0 }&0&-25&100\\& & 3& \color{orangered}{-3} & & & \\ \hline &1&-1&\color{orangered}{-3}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-4&0&0&-25&100\\& & 3& -3& \color{blue}{-9} & & \\ \hline &1&-1&\color{blue}{-3}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrrrr}3&1&-4&0&\color{orangered}{ 0 }&-25&100\\& & 3& -3& \color{orangered}{-9} & & \\ \hline &1&-1&-3&\color{orangered}{-9}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ -27 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-4&0&0&-25&100\\& & 3& -3& -9& \color{blue}{-27} & \\ \hline &1&-1&-3&\color{blue}{-9}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -25 } + \color{orangered}{ \left( -27 \right) } = \color{orangered}{ -52 } $
$$ \begin{array}{c|rrrrrr}3&1&-4&0&0&\color{orangered}{ -25 }&100\\& & 3& -3& -9& \color{orangered}{-27} & \\ \hline &1&-1&-3&-9&\color{orangered}{-52}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -52 \right) } = \color{blue}{ -156 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-4&0&0&-25&100\\& & 3& -3& -9& -27& \color{blue}{-156} \\ \hline &1&-1&-3&-9&\color{blue}{-52}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 100 } + \color{orangered}{ \left( -156 \right) } = \color{orangered}{ -56 } $
$$ \begin{array}{c|rrrrrr}3&1&-4&0&0&-25&\color{orangered}{ 100 }\\& & 3& -3& -9& -27& \color{orangered}{-156} \\ \hline &\color{blue}{1}&\color{blue}{-1}&\color{blue}{-3}&\color{blue}{-9}&\color{blue}{-52}&\color{orangered}{-56} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}-x^{3}-3x^{2}-9x-52 } $ with a remainder of $ \color{red}{ -56 } $.