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