Divide using synthetic division $$ ( x^{5}-5x^{4}+10x^{3}-10x^{2}+5x-1 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}3&1&-5&10&-10&5&-1\\& & 3& -6& 12& 6& \color{black}{33} \\ \hline &\color{blue}{1}&\color{blue}{-2}&\color{blue}{4}&\color{blue}{2}&\color{blue}{11}&\color{orangered}{32} \end{array} $$The solution is:
$$ \frac{ x^{5}-5x^{4}+10x^{3}-10x^{2}+5x-1 }{ x-3 } = \color{blue}{x^{4}-2x^{3}+4x^{2}+2x+11} ~+~ \frac{ \color{red}{ 32 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-5&10&-10&5&-1\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}3&\color{orangered}{ 1 }&-5&10&-10&5&-1\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-5&10&-10&5&-1\\& & \color{blue}{3} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 3 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrrr}3&1&\color{orangered}{ -5 }&10&-10&5&-1\\& & \color{orangered}{3} & & & & \\ \hline &1&\color{orangered}{-2}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-5&10&-10&5&-1\\& & 3& \color{blue}{-6} & & & \\ \hline &1&\color{blue}{-2}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrrr}3&1&-5&\color{orangered}{ 10 }&-10&5&-1\\& & 3& \color{orangered}{-6} & & & \\ \hline &1&-2&\color{orangered}{4}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 4 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-5&10&-10&5&-1\\& & 3& -6& \color{blue}{12} & & \\ \hline &1&-2&\color{blue}{4}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 12 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrrr}3&1&-5&10&\color{orangered}{ -10 }&5&-1\\& & 3& -6& \color{orangered}{12} & & \\ \hline &1&-2&4&\color{orangered}{2}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 2 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-5&10&-10&5&-1\\& & 3& -6& 12& \color{blue}{6} & \\ \hline &1&-2&4&\color{blue}{2}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 6 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrrrr}3&1&-5&10&-10&\color{orangered}{ 5 }&-1\\& & 3& -6& 12& \color{orangered}{6} & \\ \hline &1&-2&4&2&\color{orangered}{11}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 11 } = \color{blue}{ 33 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-5&10&-10&5&-1\\& & 3& -6& 12& 6& \color{blue}{33} \\ \hline &1&-2&4&2&\color{blue}{11}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 33 } = \color{orangered}{ 32 } $
$$ \begin{array}{c|rrrrrr}3&1&-5&10&-10&5&\color{orangered}{ -1 }\\& & 3& -6& 12& 6& \color{orangered}{33} \\ \hline &\color{blue}{1}&\color{blue}{-2}&\color{blue}{4}&\color{blue}{2}&\color{blue}{11}&\color{orangered}{32} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}-2x^{3}+4x^{2}+2x+11 } $ with a remainder of $ \color{red}{ 32 } $.