Divide using synthetic division $$ ( x^{5}-12x^{4}+6x^{3}-15x+36 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}1&1&-12&6&0&-15&36\\& & 1& -11& -5& -5& \color{black}{-20} \\ \hline &\color{blue}{1}&\color{blue}{-11}&\color{blue}{-5}&\color{blue}{-5}&\color{blue}{-20}&\color{orangered}{16} \end{array} $$The solution is:
$$ \frac{ x^{5}-12x^{4}+6x^{3}-15x+36 }{ x-1 } = \color{blue}{x^{4}-11x^{3}-5x^{2}-5x-20} ~+~ \frac{ \color{red}{ 16 } }{ x-1 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&1&-12&6&0&-15&36\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}1&\color{orangered}{ 1 }&-12&6&0&-15&36\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 1 } = \color{blue}{ 1 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&1&-12&6&0&-15&36\\& & \color{blue}{1} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 1 } = \color{orangered}{ -11 } $
$$ \begin{array}{c|rrrrrr}1&1&\color{orangered}{ -12 }&6&0&-15&36\\& & \color{orangered}{1} & & & & \\ \hline &1&\color{orangered}{-11}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -11 \right) } = \color{blue}{ -11 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&1&-12&6&0&-15&36\\& & 1& \color{blue}{-11} & & & \\ \hline &1&\color{blue}{-11}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -11 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrrr}1&1&-12&\color{orangered}{ 6 }&0&-15&36\\& & 1& \color{orangered}{-11} & & & \\ \hline &1&-11&\color{orangered}{-5}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&1&-12&6&0&-15&36\\& & 1& -11& \color{blue}{-5} & & \\ \hline &1&-11&\color{blue}{-5}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrrr}1&1&-12&6&\color{orangered}{ 0 }&-15&36\\& & 1& -11& \color{orangered}{-5} & & \\ \hline &1&-11&-5&\color{orangered}{-5}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&1&-12&6&0&-15&36\\& & 1& -11& -5& \color{blue}{-5} & \\ \hline &1&-11&-5&\color{blue}{-5}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -20 } $
$$ \begin{array}{c|rrrrrr}1&1&-12&6&0&\color{orangered}{ -15 }&36\\& & 1& -11& -5& \color{orangered}{-5} & \\ \hline &1&-11&-5&-5&\color{orangered}{-20}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -20 \right) } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&1&-12&6&0&-15&36\\& & 1& -11& -5& -5& \color{blue}{-20} \\ \hline &1&-11&-5&-5&\color{blue}{-20}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 36 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrrrrr}1&1&-12&6&0&-15&\color{orangered}{ 36 }\\& & 1& -11& -5& -5& \color{orangered}{-20} \\ \hline &\color{blue}{1}&\color{blue}{-11}&\color{blue}{-5}&\color{blue}{-5}&\color{blue}{-20}&\color{orangered}{16} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}-11x^{3}-5x^{2}-5x-20 } $ with a remainder of $ \color{red}{ 16 } $.