Find remainder when $ 2x^{4}+5x^{3}+2x^{2}-7x-15 $ is divided by $ 2x-3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&2&5&2&-7&-15\\& & 6& 33& 105& \color{black}{294} \\ \hline &\color{blue}{2}&\color{blue}{11}&\color{blue}{35}&\color{blue}{98}&\color{orangered}{279} \end{array} $$The remainder when $ 2x^{4}+5x^{3}+2x^{2}-7x-15 $ is divided by $ x-3 $ is $ \, \color{red}{ 279 } $.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&5&2&-7&-15\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 2 }&5&2&-7&-15\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&5&2&-7&-15\\& & \color{blue}{6} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 6 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrrr}3&2&\color{orangered}{ 5 }&2&-7&-15\\& & \color{orangered}{6} & & & \\ \hline &2&\color{orangered}{11}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&5&2&-7&-15\\& & 6& \color{blue}{33} & & \\ \hline &2&\color{blue}{11}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 33 } = \color{orangered}{ 35 } $
$$ \begin{array}{c|rrrrr}3&2&5&\color{orangered}{ 2 }&-7&-15\\& & 6& \color{orangered}{33} & & \\ \hline &2&11&\color{orangered}{35}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 35 } = \color{blue}{ 105 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&5&2&-7&-15\\& & 6& 33& \color{blue}{105} & \\ \hline &2&11&\color{blue}{35}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 105 } = \color{orangered}{ 98 } $
$$ \begin{array}{c|rrrrr}3&2&5&2&\color{orangered}{ -7 }&-15\\& & 6& 33& \color{orangered}{105} & \\ \hline &2&11&35&\color{orangered}{98}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 98 } = \color{blue}{ 294 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&5&2&-7&-15\\& & 6& 33& 105& \color{blue}{294} \\ \hline &2&11&35&\color{blue}{98}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 294 } = \color{orangered}{ 279 } $
$$ \begin{array}{c|rrrrr}3&2&5&2&-7&\color{orangered}{ -15 }\\& & 6& 33& 105& \color{orangered}{294} \\ \hline &\color{blue}{2}&\color{blue}{11}&\color{blue}{35}&\color{blue}{98}&\color{orangered}{279} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 279 }\right) $.