Divide using synthetic division $$ ( 3x^{4}+8x^{3}+14x^{2}+25x+10 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&3&8&14&25&10\\& & -6& -4& -20& \color{black}{-10} \\ \hline &\color{blue}{3}&\color{blue}{2}&\color{blue}{10}&\color{blue}{5}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 3x^{4}+8x^{3}+14x^{2}+25x+10 }{ x+2 } = \color{blue}{3x^{3}+2x^{2}+10x+5} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&8&14&25&10\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 3 }&8&14&25&10\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 3 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&8&14&25&10\\& & \color{blue}{-6} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}-2&3&\color{orangered}{ 8 }&14&25&10\\& & \color{orangered}{-6} & & & \\ \hline &3&\color{orangered}{2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&8&14&25&10\\& & -6& \color{blue}{-4} & & \\ \hline &3&\color{blue}{2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 14 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrr}-2&3&8&\color{orangered}{ 14 }&25&10\\& & -6& \color{orangered}{-4} & & \\ \hline &3&2&\color{orangered}{10}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 10 } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&8&14&25&10\\& & -6& -4& \color{blue}{-20} & \\ \hline &3&2&\color{blue}{10}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 25 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}-2&3&8&14&\color{orangered}{ 25 }&10\\& & -6& -4& \color{orangered}{-20} & \\ \hline &3&2&10&\color{orangered}{5}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 5 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&8&14&25&10\\& & -6& -4& -20& \color{blue}{-10} \\ \hline &3&2&10&\color{blue}{5}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-2&3&8&14&25&\color{orangered}{ 10 }\\& & -6& -4& -20& \color{orangered}{-10} \\ \hline &\color{blue}{3}&\color{blue}{2}&\color{blue}{10}&\color{blue}{5}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}+2x^{2}+10x+5 } $ with a remainder of $ \color{red}{ 0 } $.