Divide using synthetic division $$ ( x^{5}-2x^{2}+1 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-2&1&0&0&-2&0&1\\& & -2& 4& -8& 20& \color{black}{-40} \\ \hline &\color{blue}{1}&\color{blue}{-2}&\color{blue}{4}&\color{blue}{-10}&\color{blue}{20}&\color{orangered}{-39} \end{array} $$The solution is:
$$ \frac{ x^{5}-2x^{2}+1 }{ x+2 } = \color{blue}{x^{4}-2x^{3}+4x^{2}-10x+20} \color{red}{~-~} \frac{ \color{red}{ 39 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&0&0&-2&0&1\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-2&\color{orangered}{ 1 }&0&0&-2&0&1\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&0&0&-2&0&1\\& & \color{blue}{-2} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrrr}-2&1&\color{orangered}{ 0 }&0&-2&0&1\\& & \color{orangered}{-2} & & & & \\ \hline &1&\color{orangered}{-2}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&0&0&-2&0&1\\& & -2& \color{blue}{4} & & & \\ \hline &1&\color{blue}{-2}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 4 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrrr}-2&1&0&\color{orangered}{ 0 }&-2&0&1\\& & -2& \color{orangered}{4} & & & \\ \hline &1&-2&\color{orangered}{4}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 4 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&0&0&-2&0&1\\& & -2& 4& \color{blue}{-8} & & \\ \hline &1&-2&\color{blue}{4}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrrr}-2&1&0&0&\color{orangered}{ -2 }&0&1\\& & -2& 4& \color{orangered}{-8} & & \\ \hline &1&-2&4&\color{orangered}{-10}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&0&0&-2&0&1\\& & -2& 4& -8& \color{blue}{20} & \\ \hline &1&-2&4&\color{blue}{-10}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 20 } = \color{orangered}{ 20 } $
$$ \begin{array}{c|rrrrrr}-2&1&0&0&-2&\color{orangered}{ 0 }&1\\& & -2& 4& -8& \color{orangered}{20} & \\ \hline &1&-2&4&-10&\color{orangered}{20}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 20 } = \color{blue}{ -40 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&0&0&-2&0&1\\& & -2& 4& -8& 20& \color{blue}{-40} \\ \hline &1&-2&4&-10&\color{blue}{20}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -40 \right) } = \color{orangered}{ -39 } $
$$ \begin{array}{c|rrrrrr}-2&1&0&0&-2&0&\color{orangered}{ 1 }\\& & -2& 4& -8& 20& \color{orangered}{-40} \\ \hline &\color{blue}{1}&\color{blue}{-2}&\color{blue}{4}&\color{blue}{-10}&\color{blue}{20}&\color{orangered}{-39} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}-2x^{3}+4x^{2}-10x+20 } $ with a remainder of $ \color{red}{ -39 } $.