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