Find remainder when $ 2x^{4}-3x^{3}+4x^{2}+17x+7 $ is divided by $ 2x-3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&2&-3&4&17&7\\& & 6& 9& 39& \color{black}{168} \\ \hline &\color{blue}{2}&\color{blue}{3}&\color{blue}{13}&\color{blue}{56}&\color{orangered}{175} \end{array} $$The remainder when $ 2x^{4}-3x^{3}+4x^{2}+17x+7 $ is divided by $ x-3 $ is $ \, \color{red}{ 175 } $.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-3&4&17&7\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 2 }&-3&4&17&7\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 2 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-3&4&17&7\\& & \color{blue}{6} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 6 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}3&2&\color{orangered}{ -3 }&4&17&7\\& & \color{orangered}{6} & & & \\ \hline &2&\color{orangered}{3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 3 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-3&4&17&7\\& & 6& \color{blue}{9} & & \\ \hline &2&\color{blue}{3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 9 } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrrr}3&2&-3&\color{orangered}{ 4 }&17&7\\& & 6& \color{orangered}{9} & & \\ \hline &2&3&\color{orangered}{13}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 13 } = \color{blue}{ 39 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-3&4&17&7\\& & 6& 9& \color{blue}{39} & \\ \hline &2&3&\color{blue}{13}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 17 } + \color{orangered}{ 39 } = \color{orangered}{ 56 } $
$$ \begin{array}{c|rrrrr}3&2&-3&4&\color{orangered}{ 17 }&7\\& & 6& 9& \color{orangered}{39} & \\ \hline &2&3&13&\color{orangered}{56}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 56 } = \color{blue}{ 168 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-3&4&17&7\\& & 6& 9& 39& \color{blue}{168} \\ \hline &2&3&13&\color{blue}{56}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 168 } = \color{orangered}{ 175 } $
$$ \begin{array}{c|rrrrr}3&2&-3&4&17&\color{orangered}{ 7 }\\& & 6& 9& 39& \color{orangered}{168} \\ \hline &\color{blue}{2}&\color{blue}{3}&\color{blue}{13}&\color{blue}{56}&\color{orangered}{175} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 175 }\right) $.