Divide using synthetic division $$ ( x^{5}+243 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-3&1&0&0&0&0&243\\& & -3& 9& -27& 81& \color{black}{-243} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{9}&\color{blue}{-27}&\color{blue}{81}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{5}+243 }{ x+3 } = \color{blue}{x^{4}-3x^{3}+9x^{2}-27x+81} $$Explanation
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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&0&0&0&0&243\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-3&\color{orangered}{ 1 }&0&0&0&0&243\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&0&0&0&0&243\\& & \color{blue}{-3} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrrr}-3&1&\color{orangered}{ 0 }&0&0&0&243\\& & \color{orangered}{-3} & & & & \\ \hline &1&\color{orangered}{-3}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&0&0&0&0&243\\& & -3& \color{blue}{9} & & & \\ \hline &1&\color{blue}{-3}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 9 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrrr}-3&1&0&\color{orangered}{ 0 }&0&0&243\\& & -3& \color{orangered}{9} & & & \\ \hline &1&-3&\color{orangered}{9}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 9 } = \color{blue}{ -27 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&0&0&0&0&243\\& & -3& 9& \color{blue}{-27} & & \\ \hline &1&-3&\color{blue}{9}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -27 \right) } = \color{orangered}{ -27 } $
$$ \begin{array}{c|rrrrrr}-3&1&0&0&\color{orangered}{ 0 }&0&243\\& & -3& 9& \color{orangered}{-27} & & \\ \hline &1&-3&9&\color{orangered}{-27}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -27 \right) } = \color{blue}{ 81 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&0&0&0&0&243\\& & -3& 9& -27& \color{blue}{81} & \\ \hline &1&-3&9&\color{blue}{-27}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 81 } = \color{orangered}{ 81 } $
$$ \begin{array}{c|rrrrrr}-3&1&0&0&0&\color{orangered}{ 0 }&243\\& & -3& 9& -27& \color{orangered}{81} & \\ \hline &1&-3&9&-27&\color{orangered}{81}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 81 } = \color{blue}{ -243 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&0&0&0&0&243\\& & -3& 9& -27& 81& \color{blue}{-243} \\ \hline &1&-3&9&-27&\color{blue}{81}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 243 } + \color{orangered}{ \left( -243 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}-3&1&0&0&0&0&\color{orangered}{ 243 }\\& & -3& 9& -27& 81& \color{orangered}{-243} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{9}&\color{blue}{-27}&\color{blue}{81}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}-3x^{3}+9x^{2}-27x+81 } $ with a remainder of $ \color{red}{ 0 } $.