Find remainder when $ x^{4}-3x^{3}-14x^{2}+12x+40 $ is divided by $ x-3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&1&-3&-14&12&40\\& & 3& 0& -42& \color{black}{-90} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{-14}&\color{blue}{-30}&\color{orangered}{-50} \end{array} $$The remainder when $ x^{4}-3x^{3}-14x^{2}+12x+40 $ is divided by $ x-3 $ is $ \, \color{red}{ -50 } $.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-3&-14&12&40\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 1 }&-3&-14&12&40\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-3&-14&12&40\\& & \color{blue}{3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 3 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}3&1&\color{orangered}{ -3 }&-14&12&40\\& & \color{orangered}{3} & & & \\ \hline &1&\color{orangered}{0}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-3&-14&12&40\\& & 3& \color{blue}{0} & & \\ \hline &1&\color{blue}{0}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 0 } = \color{orangered}{ -14 } $
$$ \begin{array}{c|rrrrr}3&1&-3&\color{orangered}{ -14 }&12&40\\& & 3& \color{orangered}{0} & & \\ \hline &1&0&\color{orangered}{-14}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -14 \right) } = \color{blue}{ -42 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-3&-14&12&40\\& & 3& 0& \color{blue}{-42} & \\ \hline &1&0&\color{blue}{-14}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -42 \right) } = \color{orangered}{ -30 } $
$$ \begin{array}{c|rrrrr}3&1&-3&-14&\color{orangered}{ 12 }&40\\& & 3& 0& \color{orangered}{-42} & \\ \hline &1&0&-14&\color{orangered}{-30}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -30 \right) } = \color{blue}{ -90 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-3&-14&12&40\\& & 3& 0& -42& \color{blue}{-90} \\ \hline &1&0&-14&\color{blue}{-30}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 40 } + \color{orangered}{ \left( -90 \right) } = \color{orangered}{ -50 } $
$$ \begin{array}{c|rrrrr}3&1&-3&-14&12&\color{orangered}{ 40 }\\& & 3& 0& -42& \color{orangered}{-90} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{-14}&\color{blue}{-30}&\color{orangered}{-50} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -50 }\right) $.