Divide using synthetic division $$ ( x^{4}-5x+3 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&1&0&0&-5&3\\& & -3& 9& -27& \color{black}{96} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{9}&\color{blue}{-32}&\color{orangered}{99} \end{array} $$The solution is:
$$ \frac{ x^{4}-5x+3 }{ x+3 } = \color{blue}{x^{3}-3x^{2}+9x-32} ~+~ \frac{ \color{red}{ 99 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&0&0&-5&3\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 1 }&0&0&-5&3\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 1 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&0&0&-5&3\\& & \color{blue}{-3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}-3&1&\color{orangered}{ 0 }&0&-5&3\\& & \color{orangered}{-3} & & & \\ \hline &1&\color{orangered}{-3}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&0&0&-5&3\\& & -3& \color{blue}{9} & & \\ \hline &1&\color{blue}{-3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 9 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrr}-3&1&0&\color{orangered}{ 0 }&-5&3\\& & -3& \color{orangered}{9} & & \\ \hline &1&-3&\color{orangered}{9}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&0&0&-5&3\\& & -3& 9& \color{blue}{-27} & \\ \hline &1&-3&\color{blue}{9}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ \left( -27 \right) } = \color{orangered}{ -32 } $
$$ \begin{array}{c|rrrrr}-3&1&0&0&\color{orangered}{ -5 }&3\\& & -3& 9& \color{orangered}{-27} & \\ \hline &1&-3&9&\color{orangered}{-32}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -32 \right) } = \color{blue}{ 96 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&0&0&-5&3\\& & -3& 9& -27& \color{blue}{96} \\ \hline &1&-3&9&\color{blue}{-32}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 96 } = \color{orangered}{ 99 } $
$$ \begin{array}{c|rrrrr}-3&1&0&0&-5&\color{orangered}{ 3 }\\& & -3& 9& -27& \color{orangered}{96} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{9}&\color{blue}{-32}&\color{orangered}{99} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-3x^{2}+9x-32 } $ with a remainder of $ \color{red}{ 99 } $.