Find remainder when $ x^{4}-3x^{3}-4x^{2}-6x+15 $ is divided by $ x-1 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&1&-3&-4&-6&15\\& & 1& -2& -6& \color{black}{-12} \\ \hline &\color{blue}{1}&\color{blue}{-2}&\color{blue}{-6}&\color{blue}{-12}&\color{orangered}{3} \end{array} $$The remainder when $ x^{4}-3x^{3}-4x^{2}-6x+15 $ is divided by $ x-1 $ is $ \, \color{red}{ 3 } $.
Explanation
We can find remainder using synthetic division method.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&-3&-4&-6&15\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 1 }&-3&-4&-6&15\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&-3&-4&-6&15\\& & \color{blue}{1} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 1 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}1&1&\color{orangered}{ -3 }&-4&-6&15\\& & \color{orangered}{1} & & & \\ \hline &1&\color{orangered}{-2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&-3&-4&-6&15\\& & 1& \color{blue}{-2} & & \\ \hline &1&\color{blue}{-2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}1&1&-3&\color{orangered}{ -4 }&-6&15\\& & 1& \color{orangered}{-2} & & \\ \hline &1&-2&\color{orangered}{-6}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&-3&-4&-6&15\\& & 1& -2& \color{blue}{-6} & \\ \hline &1&-2&\color{blue}{-6}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrrr}1&1&-3&-4&\color{orangered}{ -6 }&15\\& & 1& -2& \color{orangered}{-6} & \\ \hline &1&-2&-6&\color{orangered}{-12}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&-3&-4&-6&15\\& & 1& -2& -6& \color{blue}{-12} \\ \hline &1&-2&-6&\color{blue}{-12}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}1&1&-3&-4&-6&\color{orangered}{ 15 }\\& & 1& -2& -6& \color{orangered}{-12} \\ \hline &\color{blue}{1}&\color{blue}{-2}&\color{blue}{-6}&\color{blue}{-12}&\color{orangered}{3} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 3 }\right) $.