Divide using synthetic division $$ ( 5x^{4}+12x^{3}+2x^{2}+10 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&5&12&2&0&10\\& & -10& -4& 4& \color{black}{-8} \\ \hline &\color{blue}{5}&\color{blue}{2}&\color{blue}{-2}&\color{blue}{4}&\color{orangered}{2} \end{array} $$The solution is:
$$ \frac{ 5x^{4}+12x^{3}+2x^{2}+10 }{ x+2 } = \color{blue}{5x^{3}+2x^{2}-2x+4} ~+~ \frac{ \color{red}{ 2 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&5&12&2&0&10\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 5 }&12&2&0&10\\& & & & & \\ \hline &\color{orangered}{5}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&5&12&2&0&10\\& & \color{blue}{-10} & & & \\ \hline &\color{blue}{5}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}-2&5&\color{orangered}{ 12 }&2&0&10\\& & \color{orangered}{-10} & & & \\ \hline &5&\color{orangered}{2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 2 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&5&12&2&0&10\\& & -10& \color{blue}{-4} & & \\ \hline &5&\color{blue}{2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}-2&5&12&\color{orangered}{ 2 }&0&10\\& & -10& \color{orangered}{-4} & & \\ \hline &5&2&\color{orangered}{-2}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&5&12&2&0&10\\& & -10& -4& \color{blue}{4} & \\ \hline &5&2&\color{blue}{-2}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 4 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}-2&5&12&2&\color{orangered}{ 0 }&10\\& & -10& -4& \color{orangered}{4} & \\ \hline &5&2&-2&\color{orangered}{4}& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&5&12&2&0&10\\& & -10& -4& 4& \color{blue}{-8} \\ \hline &5&2&-2&\color{blue}{4}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}-2&5&12&2&0&\color{orangered}{ 10 }\\& & -10& -4& 4& \color{orangered}{-8} \\ \hline &\color{blue}{5}&\color{blue}{2}&\color{blue}{-2}&\color{blue}{4}&\color{orangered}{2} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{3}+2x^{2}-2x+4 } $ with a remainder of $ \color{red}{ 2 } $.