Find remainder when $ 2x^{5}+5x^{4}+12x^{2}+15x-2 $ is divided by $ x-3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}3&2&5&0&12&15&-2\\& & 6& 33& 99& 333& \color{black}{1044} \\ \hline &\color{blue}{2}&\color{blue}{11}&\color{blue}{33}&\color{blue}{111}&\color{blue}{348}&\color{orangered}{1042} \end{array} $$The remainder when $ 2x^{5}+5x^{4}+12x^{2}+15x-2 $ is divided by $ x-3 $ is $ \, \color{red}{ 1042 } $.
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.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&2&5&0&12&15&-2\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}3&\color{orangered}{ 2 }&5&0&12&15&-2\\& & & & & & \\ \hline &\color{orangered}{2}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&2&5&0&12&15&-2\\& & \color{blue}{6} & & & & \\ \hline &\color{blue}{2}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 6 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrrrr}3&2&\color{orangered}{ 5 }&0&12&15&-2\\& & \color{orangered}{6} & & & & \\ \hline &2&\color{orangered}{11}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 11 } = \color{blue}{ 33 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&2&5&0&12&15&-2\\& & 6& \color{blue}{33} & & & \\ \hline &2&\color{blue}{11}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 33 } = \color{orangered}{ 33 } $
$$ \begin{array}{c|rrrrrr}3&2&5&\color{orangered}{ 0 }&12&15&-2\\& & 6& \color{orangered}{33} & & & \\ \hline &2&11&\color{orangered}{33}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 33 } = \color{blue}{ 99 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&2&5&0&12&15&-2\\& & 6& 33& \color{blue}{99} & & \\ \hline &2&11&\color{blue}{33}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 99 } = \color{orangered}{ 111 } $
$$ \begin{array}{c|rrrrrr}3&2&5&0&\color{orangered}{ 12 }&15&-2\\& & 6& 33& \color{orangered}{99} & & \\ \hline &2&11&33&\color{orangered}{111}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 111 } = \color{blue}{ 333 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&2&5&0&12&15&-2\\& & 6& 33& 99& \color{blue}{333} & \\ \hline &2&11&33&\color{blue}{111}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ 333 } = \color{orangered}{ 348 } $
$$ \begin{array}{c|rrrrrr}3&2&5&0&12&\color{orangered}{ 15 }&-2\\& & 6& 33& 99& \color{orangered}{333} & \\ \hline &2&11&33&111&\color{orangered}{348}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 348 } = \color{blue}{ 1044 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&2&5&0&12&15&-2\\& & 6& 33& 99& 333& \color{blue}{1044} \\ \hline &2&11&33&111&\color{blue}{348}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 1044 } = \color{orangered}{ 1042 } $
$$ \begin{array}{c|rrrrrr}3&2&5&0&12&15&\color{orangered}{ -2 }\\& & 6& 33& 99& 333& \color{orangered}{1044} \\ \hline &\color{blue}{2}&\color{blue}{11}&\color{blue}{33}&\color{blue}{111}&\color{blue}{348}&\color{orangered}{1042} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 1042 }\right) $.