Divide using synthetic division $$ ( x^{3}-4x^{2}+6x-4 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}5&1&-4&6&-4\\& & 5& 5& \color{black}{55} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{11}&\color{orangered}{51} \end{array} $$The solution is:
$$ \frac{ x^{3}-4x^{2}+6x-4 }{ x-5 } = \color{blue}{x^{2}+x+11} ~+~ \frac{ \color{red}{ 51 } }{ x-5 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&-4&6&-4\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}5&\color{orangered}{ 1 }&-4&6&-4\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&-4&6&-4\\& & \color{blue}{5} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 5 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrr}5&1&\color{orangered}{ -4 }&6&-4\\& & \color{orangered}{5} & & \\ \hline &1&\color{orangered}{1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 1 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&-4&6&-4\\& & 5& \color{blue}{5} & \\ \hline &1&\color{blue}{1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 5 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrr}5&1&-4&\color{orangered}{ 6 }&-4\\& & 5& \color{orangered}{5} & \\ \hline &1&1&\color{orangered}{11}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 11 } = \color{blue}{ 55 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&-4&6&-4\\& & 5& 5& \color{blue}{55} \\ \hline &1&1&\color{blue}{11}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 55 } = \color{orangered}{ 51 } $
$$ \begin{array}{c|rrrr}5&1&-4&6&\color{orangered}{ -4 }\\& & 5& 5& \color{orangered}{55} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{11}&\color{orangered}{51} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+x+11 } $ with a remainder of $ \color{red}{ 51 } $.