Find remainder when $ x^{5}-17x^{3}+36 $ is divided by $ x-2 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}2&1&0&-17&0&0&36\\& & 2& 4& -26& -52& \color{black}{-104} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{-13}&\color{blue}{-26}&\color{blue}{-52}&\color{orangered}{-68} \end{array} $$The remainder when $ x^{5}-17x^{3}+36 $ is divided by $ x-2 $ is $ \, \color{red}{ -68 } $.
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 -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&0&-17&0&0&36\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}2&\color{orangered}{ 1 }&0&-17&0&0&36\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 1 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&0&-17&0&0&36\\& & \color{blue}{2} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 2 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrrr}2&1&\color{orangered}{ 0 }&-17&0&0&36\\& & \color{orangered}{2} & & & & \\ \hline &1&\color{orangered}{2}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 2 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&0&-17&0&0&36\\& & 2& \color{blue}{4} & & & \\ \hline &1&\color{blue}{2}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -17 } + \color{orangered}{ 4 } = \color{orangered}{ -13 } $
$$ \begin{array}{c|rrrrrr}2&1&0&\color{orangered}{ -17 }&0&0&36\\& & 2& \color{orangered}{4} & & & \\ \hline &1&2&\color{orangered}{-13}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -13 \right) } = \color{blue}{ -26 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&0&-17&0&0&36\\& & 2& 4& \color{blue}{-26} & & \\ \hline &1&2&\color{blue}{-13}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -26 \right) } = \color{orangered}{ -26 } $
$$ \begin{array}{c|rrrrrr}2&1&0&-17&\color{orangered}{ 0 }&0&36\\& & 2& 4& \color{orangered}{-26} & & \\ \hline &1&2&-13&\color{orangered}{-26}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -26 \right) } = \color{blue}{ -52 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&0&-17&0&0&36\\& & 2& 4& -26& \color{blue}{-52} & \\ \hline &1&2&-13&\color{blue}{-26}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -52 \right) } = \color{orangered}{ -52 } $
$$ \begin{array}{c|rrrrrr}2&1&0&-17&0&\color{orangered}{ 0 }&36\\& & 2& 4& -26& \color{orangered}{-52} & \\ \hline &1&2&-13&-26&\color{orangered}{-52}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -52 \right) } = \color{blue}{ -104 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&0&-17&0&0&36\\& & 2& 4& -26& -52& \color{blue}{-104} \\ \hline &1&2&-13&-26&\color{blue}{-52}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 36 } + \color{orangered}{ \left( -104 \right) } = \color{orangered}{ -68 } $
$$ \begin{array}{c|rrrrrr}2&1&0&-17&0&0&\color{orangered}{ 36 }\\& & 2& 4& -26& -52& \color{orangered}{-104} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{-13}&\color{blue}{-26}&\color{blue}{-52}&\color{orangered}{-68} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -68 }\right) $.