Find remainder when $ x^{3}+4x^{2}-5x+3 $ is divided by $ x+2 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&1&4&-5&3\\& & -2& -4& \color{black}{18} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{-9}&\color{orangered}{21} \end{array} $$The remainder when $ x^{3}+4x^{2}-5x+3 $ is divided by $ x+2 $ is $ \, \color{red}{ 21 } $.
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 + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&4&-5&3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 1 }&4&-5&3\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 1 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&4&-5&3\\& & \color{blue}{-2} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}-2&1&\color{orangered}{ 4 }&-5&3\\& & \color{orangered}{-2} & & \\ \hline &1&\color{orangered}{2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 2 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&4&-5&3\\& & -2& \color{blue}{-4} & \\ \hline &1&\color{blue}{2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrr}-2&1&4&\color{orangered}{ -5 }&3\\& & -2& \color{orangered}{-4} & \\ \hline &1&2&\color{orangered}{-9}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&4&-5&3\\& & -2& -4& \color{blue}{18} \\ \hline &1&2&\color{blue}{-9}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 18 } = \color{orangered}{ 21 } $
$$ \begin{array}{c|rrrr}-2&1&4&-5&\color{orangered}{ 3 }\\& & -2& -4& \color{orangered}{18} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{-9}&\color{orangered}{21} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 21 }\right) $.