Divide using synthetic division $$ ( x^{4}-4x^{3}+22x^{2}+28x-203 ) \div ( -5x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&1&-4&22&28&-203\\& & -2& 12& -68& \color{black}{80} \\ \hline &\color{blue}{1}&\color{blue}{-6}&\color{blue}{34}&\color{blue}{-40}&\color{orangered}{-123} \end{array} $$The solution is:
$$ \frac{ x^{4}-4x^{3}+22x^{2}+28x-203 }{ x+2 } = \color{blue}{x^{3}-6x^{2}+34x-40} \color{red}{~-~} \frac{ \color{red}{ 123 } }{ x+2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-4&22&28&-203\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 1 }&-4&22&28&-203\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 1 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-4&22&28&-203\\& & \color{blue}{-2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}-2&1&\color{orangered}{ -4 }&22&28&-203\\& & \color{orangered}{-2} & & & \\ \hline &1&\color{orangered}{-6}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-4&22&28&-203\\& & -2& \color{blue}{12} & & \\ \hline &1&\color{blue}{-6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 22 } + \color{orangered}{ 12 } = \color{orangered}{ 34 } $
$$ \begin{array}{c|rrrrr}-2&1&-4&\color{orangered}{ 22 }&28&-203\\& & -2& \color{orangered}{12} & & \\ \hline &1&-6&\color{orangered}{34}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 34 } = \color{blue}{ -68 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-4&22&28&-203\\& & -2& 12& \color{blue}{-68} & \\ \hline &1&-6&\color{blue}{34}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 28 } + \color{orangered}{ \left( -68 \right) } = \color{orangered}{ -40 } $
$$ \begin{array}{c|rrrrr}-2&1&-4&22&\color{orangered}{ 28 }&-203\\& & -2& 12& \color{orangered}{-68} & \\ \hline &1&-6&34&\color{orangered}{-40}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -40 \right) } = \color{blue}{ 80 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-4&22&28&-203\\& & -2& 12& -68& \color{blue}{80} \\ \hline &1&-6&34&\color{blue}{-40}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -203 } + \color{orangered}{ 80 } = \color{orangered}{ -123 } $
$$ \begin{array}{c|rrrrr}-2&1&-4&22&28&\color{orangered}{ -203 }\\& & -2& 12& -68& \color{orangered}{80} \\ \hline &\color{blue}{1}&\color{blue}{-6}&\color{blue}{34}&\color{blue}{-40}&\color{orangered}{-123} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-6x^{2}+34x-40 } $ with a remainder of $ \color{red}{ -123 } $.