Find remainder when $ 4x^{3}+11x^{2}+14 $ is divided by $ x+3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&4&11&0&14\\& & -12& 3& \color{black}{-9} \\ \hline &\color{blue}{4}&\color{blue}{-1}&\color{blue}{3}&\color{orangered}{5} \end{array} $$The remainder when $ 4x^{3}+11x^{2}+14 $ is divided by $ x+3 $ is $ \, \color{red}{ 5 } $.
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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&11&0&14\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 4 }&11&0&14\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 4 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&11&0&14\\& & \color{blue}{-12} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}-3&4&\color{orangered}{ 11 }&0&14\\& & \color{orangered}{-12} & & \\ \hline &4&\color{orangered}{-1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&11&0&14\\& & -12& \color{blue}{3} & \\ \hline &4&\color{blue}{-1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 3 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}-3&4&11&\color{orangered}{ 0 }&14\\& & -12& \color{orangered}{3} & \\ \hline &4&-1&\color{orangered}{3}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 3 } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&11&0&14\\& & -12& 3& \color{blue}{-9} \\ \hline &4&-1&\color{blue}{3}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 14 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}-3&4&11&0&\color{orangered}{ 14 }\\& & -12& 3& \color{orangered}{-9} \\ \hline &\color{blue}{4}&\color{blue}{-1}&\color{blue}{3}&\color{orangered}{5} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 5 }\right) $.