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