Find remainder when $ 2x^{4}+5 $ is divided by $ x-4 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}4&2&0&0&0&5\\& & 8& 32& 128& \color{black}{512} \\ \hline &\color{blue}{2}&\color{blue}{8}&\color{blue}{32}&\color{blue}{128}&\color{orangered}{517} \end{array} $$The remainder when $ 2x^{4}+5 $ is divided by $ x-4 $ is $ \, \color{red}{ 517 } $.
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 -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&0&0&0&5\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}4&\color{orangered}{ 2 }&0&0&0&5\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 2 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&0&0&0&5\\& & \color{blue}{8} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 8 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}4&2&\color{orangered}{ 0 }&0&0&5\\& & \color{orangered}{8} & & & \\ \hline &2&\color{orangered}{8}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 8 } = \color{blue}{ 32 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&0&0&0&5\\& & 8& \color{blue}{32} & & \\ \hline &2&\color{blue}{8}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 32 } = \color{orangered}{ 32 } $
$$ \begin{array}{c|rrrrr}4&2&0&\color{orangered}{ 0 }&0&5\\& & 8& \color{orangered}{32} & & \\ \hline &2&8&\color{orangered}{32}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 32 } = \color{blue}{ 128 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&0&0&0&5\\& & 8& 32& \color{blue}{128} & \\ \hline &2&8&\color{blue}{32}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 128 } = \color{orangered}{ 128 } $
$$ \begin{array}{c|rrrrr}4&2&0&0&\color{orangered}{ 0 }&5\\& & 8& 32& \color{orangered}{128} & \\ \hline &2&8&32&\color{orangered}{128}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 128 } = \color{blue}{ 512 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&0&0&0&5\\& & 8& 32& 128& \color{blue}{512} \\ \hline &2&8&32&\color{blue}{128}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 512 } = \color{orangered}{ 517 } $
$$ \begin{array}{c|rrrrr}4&2&0&0&0&\color{orangered}{ 5 }\\& & 8& 32& 128& \color{orangered}{512} \\ \hline &\color{blue}{2}&\color{blue}{8}&\color{blue}{32}&\color{blue}{128}&\color{orangered}{517} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 517 }\right) $.