Find remainder when $ x^{4}-3x^{3}-14x^{2}+12x+40 $ is divided by $ x+5 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-5&1&-3&-14&12&40\\& & -5& 40& -130& \color{black}{590} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{blue}{26}&\color{blue}{-118}&\color{orangered}{630} \end{array} $$The remainder when $ x^{4}-3x^{3}-14x^{2}+12x+40 $ is divided by $ x+5 $ is $ \, \color{red}{ 630 } $.
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 + 5 = 0 $ ( $ x = \color{blue}{ -5 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&1&-3&-14&12&40\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-5&\color{orangered}{ 1 }&-3&-14&12&40\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 1 } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&1&-3&-14&12&40\\& & \color{blue}{-5} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}-5&1&\color{orangered}{ -3 }&-14&12&40\\& & \color{orangered}{-5} & & & \\ \hline &1&\color{orangered}{-8}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 40 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&1&-3&-14&12&40\\& & -5& \color{blue}{40} & & \\ \hline &1&\color{blue}{-8}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 40 } = \color{orangered}{ 26 } $
$$ \begin{array}{c|rrrrr}-5&1&-3&\color{orangered}{ -14 }&12&40\\& & -5& \color{orangered}{40} & & \\ \hline &1&-8&\color{orangered}{26}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 26 } = \color{blue}{ -130 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&1&-3&-14&12&40\\& & -5& 40& \color{blue}{-130} & \\ \hline &1&-8&\color{blue}{26}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -130 \right) } = \color{orangered}{ -118 } $
$$ \begin{array}{c|rrrrr}-5&1&-3&-14&\color{orangered}{ 12 }&40\\& & -5& 40& \color{orangered}{-130} & \\ \hline &1&-8&26&\color{orangered}{-118}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -118 \right) } = \color{blue}{ 590 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&1&-3&-14&12&40\\& & -5& 40& -130& \color{blue}{590} \\ \hline &1&-8&26&\color{blue}{-118}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 40 } + \color{orangered}{ 590 } = \color{orangered}{ 630 } $
$$ \begin{array}{c|rrrrr}-5&1&-3&-14&12&\color{orangered}{ 40 }\\& & -5& 40& -130& \color{orangered}{590} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{blue}{26}&\color{blue}{-118}&\color{orangered}{630} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 630 }\right) $.