Find remainder when $ 2x^{3}-296x^{2}+7x+5 $ is divided by $ x-3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&2&-296&7&5\\& & 6& -870& \color{black}{-2589} \\ \hline &\color{blue}{2}&\color{blue}{-290}&\color{blue}{-863}&\color{orangered}{-2584} \end{array} $$The remainder when $ 2x^{3}-296x^{2}+7x+5 $ is divided by $ x-3 $ is $ \, \color{red}{ -2584 } $.
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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&2&-296&7&5\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 2 }&-296&7&5\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 2 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&2&-296&7&5\\& & \color{blue}{6} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -296 } + \color{orangered}{ 6 } = \color{orangered}{ -290 } $
$$ \begin{array}{c|rrrr}3&2&\color{orangered}{ -296 }&7&5\\& & \color{orangered}{6} & & \\ \hline &2&\color{orangered}{-290}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -290 \right) } = \color{blue}{ -870 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&2&-296&7&5\\& & 6& \color{blue}{-870} & \\ \hline &2&\color{blue}{-290}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -870 \right) } = \color{orangered}{ -863 } $
$$ \begin{array}{c|rrrr}3&2&-296&\color{orangered}{ 7 }&5\\& & 6& \color{orangered}{-870} & \\ \hline &2&-290&\color{orangered}{-863}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -863 \right) } = \color{blue}{ -2589 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&2&-296&7&5\\& & 6& -870& \color{blue}{-2589} \\ \hline &2&-290&\color{blue}{-863}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -2589 \right) } = \color{orangered}{ -2584 } $
$$ \begin{array}{c|rrrr}3&2&-296&7&\color{orangered}{ 5 }\\& & 6& -870& \color{orangered}{-2589} \\ \hline &\color{blue}{2}&\color{blue}{-290}&\color{blue}{-863}&\color{orangered}{-2584} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -2584 }\right) $.