Find remainder when $ 4x^{3}+11x^{2}+30x+18 $ is divided by $ 4x+3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&4&11&30&18\\& & -12& 3& \color{black}{-99} \\ \hline &\color{blue}{4}&\color{blue}{-1}&\color{blue}{33}&\color{orangered}{-81} \end{array} $$The remainder when $ 4x^{3}+11x^{2}+30x+18 $ is divided by $ x+3 $ is $ \, \color{red}{ -81 } $.
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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&11&30&18\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 4 }&11&30&18\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&11&30&18\\& & \color{blue}{-12} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}-3&4&\color{orangered}{ 11 }&30&18\\& & \color{orangered}{-12} & & \\ \hline &4&\color{orangered}{-1}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&11&30&18\\& & -12& \color{blue}{3} & \\ \hline &4&\color{blue}{-1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 30 } + \color{orangered}{ 3 } = \color{orangered}{ 33 } $
$$ \begin{array}{c|rrrr}-3&4&11&\color{orangered}{ 30 }&18\\& & -12& \color{orangered}{3} & \\ \hline &4&-1&\color{orangered}{33}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 33 } = \color{blue}{ -99 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&11&30&18\\& & -12& 3& \color{blue}{-99} \\ \hline &4&-1&\color{blue}{33}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 18 } + \color{orangered}{ \left( -99 \right) } = \color{orangered}{ -81 } $
$$ \begin{array}{c|rrrr}-3&4&11&30&\color{orangered}{ 18 }\\& & -12& 3& \color{orangered}{-99} \\ \hline &\color{blue}{4}&\color{blue}{-1}&\color{blue}{33}&\color{orangered}{-81} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -81 }\right) $.