Find remainder when $ x^{4}+5x^{3}-6x+3 $ is divided by $ x+3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&1&5&0&-6&3\\& & -3& -6& 18& \color{black}{-36} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{-6}&\color{blue}{12}&\color{orangered}{-33} \end{array} $$The remainder when $ x^{4}+5x^{3}-6x+3 $ is divided by $ x+3 $ is $ \, \color{red}{ -33 } $.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&5&0&-6&3\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 1 }&5&0&-6&3\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 1 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&5&0&-6&3\\& & \color{blue}{-3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}-3&1&\color{orangered}{ 5 }&0&-6&3\\& & \color{orangered}{-3} & & & \\ \hline &1&\color{orangered}{2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 2 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&5&0&-6&3\\& & -3& \color{blue}{-6} & & \\ \hline &1&\color{blue}{2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}-3&1&5&\color{orangered}{ 0 }&-6&3\\& & -3& \color{orangered}{-6} & & \\ \hline &1&2&\color{orangered}{-6}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&5&0&-6&3\\& & -3& -6& \color{blue}{18} & \\ \hline &1&2&\color{blue}{-6}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 18 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrr}-3&1&5&0&\color{orangered}{ -6 }&3\\& & -3& -6& \color{orangered}{18} & \\ \hline &1&2&-6&\color{orangered}{12}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 12 } = \color{blue}{ -36 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&5&0&-6&3\\& & -3& -6& 18& \color{blue}{-36} \\ \hline &1&2&-6&\color{blue}{12}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -36 \right) } = \color{orangered}{ -33 } $
$$ \begin{array}{c|rrrrr}-3&1&5&0&-6&\color{orangered}{ 3 }\\& & -3& -6& 18& \color{orangered}{-36} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{-6}&\color{blue}{12}&\color{orangered}{-33} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -33 }\right) $.