Find remainder when $ x^{3}-x^{2}-53 $ is divided by $ x-3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&1&-1&0&-53\\& & 3& 6& \color{black}{18} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{6}&\color{orangered}{-35} \end{array} $$The remainder when $ x^{3}-x^{2}-53 $ is divided by $ x-3 $ is $ \, \color{red}{ -35 } $.
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 -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-1&0&-53\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 1 }&-1&0&-53\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-1&0&-53\\& & \color{blue}{3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 3 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}3&1&\color{orangered}{ -1 }&0&-53\\& & \color{orangered}{3} & & \\ \hline &1&\color{orangered}{2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 2 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-1&0&-53\\& & 3& \color{blue}{6} & \\ \hline &1&\color{blue}{2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 6 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrr}3&1&-1&\color{orangered}{ 0 }&-53\\& & 3& \color{orangered}{6} & \\ \hline &1&2&\color{orangered}{6}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 6 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-1&0&-53\\& & 3& 6& \color{blue}{18} \\ \hline &1&2&\color{blue}{6}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -53 } + \color{orangered}{ 18 } = \color{orangered}{ -35 } $
$$ \begin{array}{c|rrrr}3&1&-1&0&\color{orangered}{ -53 }\\& & 3& 6& \color{orangered}{18} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{6}&\color{orangered}{-35} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -35 }\right) $.