Find remainder when $ x^{4}+x^{3}+x^{2}-63 $ is divided by $ x+3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&1&1&1&0&-63\\& & -3& 6& -21& \color{black}{63} \\ \hline &\color{blue}{1}&\color{blue}{-2}&\color{blue}{7}&\color{blue}{-21}&\color{orangered}{0} \end{array} $$The remainder when $ x^{4}+x^{3}+x^{2}-63 $ is divided by $ x+3 $ is $ \, \color{red}{ 0 } $.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&1&1&0&-63\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 1 }&1&1&0&-63\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&1&1&0&-63\\& & \color{blue}{-3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}-3&1&\color{orangered}{ 1 }&1&0&-63\\& & \color{orangered}{-3} & & & \\ \hline &1&\color{orangered}{-2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&1&1&0&-63\\& & -3& \color{blue}{6} & & \\ \hline &1&\color{blue}{-2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 6 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrr}-3&1&1&\color{orangered}{ 1 }&0&-63\\& & -3& \color{orangered}{6} & & \\ \hline &1&-2&\color{orangered}{7}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 7 } = \color{blue}{ -21 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&1&1&0&-63\\& & -3& 6& \color{blue}{-21} & \\ \hline &1&-2&\color{blue}{7}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -21 \right) } = \color{orangered}{ -21 } $
$$ \begin{array}{c|rrrrr}-3&1&1&1&\color{orangered}{ 0 }&-63\\& & -3& 6& \color{orangered}{-21} & \\ \hline &1&-2&7&\color{orangered}{-21}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -21 \right) } = \color{blue}{ 63 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&1&1&0&-63\\& & -3& 6& -21& \color{blue}{63} \\ \hline &1&-2&7&\color{blue}{-21}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -63 } + \color{orangered}{ 63 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-3&1&1&1&0&\color{orangered}{ -63 }\\& & -3& 6& -21& \color{orangered}{63} \\ \hline &\color{blue}{1}&\color{blue}{-2}&\color{blue}{7}&\color{blue}{-21}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right) $.