Divide using synthetic division $$ ( 4x^{4}-2x^{3}-17x^{2}+14x+10 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&4&-2&-17&14&10\\& & -8& 20& -6& \color{black}{-16} \\ \hline &\color{blue}{4}&\color{blue}{-10}&\color{blue}{3}&\color{blue}{8}&\color{orangered}{-6} \end{array} $$The solution is:
$$ \frac{ 4x^{4}-2x^{3}-17x^{2}+14x+10 }{ x+2 } = \color{blue}{4x^{3}-10x^{2}+3x+8} \color{red}{~-~} \frac{ \color{red}{ 6 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&4&-2&-17&14&10\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 4 }&-2&-17&14&10\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 4 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&4&-2&-17&14&10\\& & \color{blue}{-8} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrr}-2&4&\color{orangered}{ -2 }&-17&14&10\\& & \color{orangered}{-8} & & & \\ \hline &4&\color{orangered}{-10}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&4&-2&-17&14&10\\& & -8& \color{blue}{20} & & \\ \hline &4&\color{blue}{-10}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -17 } + \color{orangered}{ 20 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}-2&4&-2&\color{orangered}{ -17 }&14&10\\& & -8& \color{orangered}{20} & & \\ \hline &4&-10&\color{orangered}{3}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 3 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&4&-2&-17&14&10\\& & -8& 20& \color{blue}{-6} & \\ \hline &4&-10&\color{blue}{3}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 14 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}-2&4&-2&-17&\color{orangered}{ 14 }&10\\& & -8& 20& \color{orangered}{-6} & \\ \hline &4&-10&3&\color{orangered}{8}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 8 } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&4&-2&-17&14&10\\& & -8& 20& -6& \color{blue}{-16} \\ \hline &4&-10&3&\color{blue}{8}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}-2&4&-2&-17&14&\color{orangered}{ 10 }\\& & -8& 20& -6& \color{orangered}{-16} \\ \hline &\color{blue}{4}&\color{blue}{-10}&\color{blue}{3}&\color{blue}{8}&\color{orangered}{-6} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}-10x^{2}+3x+8 } $ with a remainder of $ \color{red}{ -6 } $.