Find remainder when $ 11x^{3}-x^{2}-143x+13 $ is divided by $ x-1 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&11&-1&-143&13\\& & 11& 10& \color{black}{-133} \\ \hline &\color{blue}{11}&\color{blue}{10}&\color{blue}{-133}&\color{orangered}{-120} \end{array} $$The remainder when $ 11x^{3}-x^{2}-143x+13 $ is divided by $ x-1 $ is $ \, \color{red}{ -120 } $.
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.
$$ \begin{array}{c|rrrr}\color{blue}{1}&11&-1&-143&13\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 11 }&-1&-143&13\\& & & & \\ \hline &\color{orangered}{11}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 11 } = \color{blue}{ 11 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&11&-1&-143&13\\& & \color{blue}{11} & & \\ \hline &\color{blue}{11}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 11 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrr}1&11&\color{orangered}{ -1 }&-143&13\\& & \color{orangered}{11} & & \\ \hline &11&\color{orangered}{10}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 10 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&11&-1&-143&13\\& & 11& \color{blue}{10} & \\ \hline &11&\color{blue}{10}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -143 } + \color{orangered}{ 10 } = \color{orangered}{ -133 } $
$$ \begin{array}{c|rrrr}1&11&-1&\color{orangered}{ -143 }&13\\& & 11& \color{orangered}{10} & \\ \hline &11&10&\color{orangered}{-133}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -133 \right) } = \color{blue}{ -133 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&11&-1&-143&13\\& & 11& 10& \color{blue}{-133} \\ \hline &11&10&\color{blue}{-133}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 13 } + \color{orangered}{ \left( -133 \right) } = \color{orangered}{ -120 } $
$$ \begin{array}{c|rrrr}1&11&-1&-143&\color{orangered}{ 13 }\\& & 11& 10& \color{orangered}{-133} \\ \hline &\color{blue}{11}&\color{blue}{10}&\color{blue}{-133}&\color{orangered}{-120} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -120 }\right) $.