Divide using synthetic division $$ ( 2x^{4}+10x^{3}+8x^{2}+x+24 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&2&10&8&1&24\\& & -4& -12& 8& \color{black}{-18} \\ \hline &\color{blue}{2}&\color{blue}{6}&\color{blue}{-4}&\color{blue}{9}&\color{orangered}{6} \end{array} $$The solution is:
$$ \frac{ 2x^{4}+10x^{3}+8x^{2}+x+24 }{ x+2 } = \color{blue}{2x^{3}+6x^{2}-4x+9} ~+~ \frac{ \color{red}{ 6 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&2&10&8&1&24\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 2 }&10&8&1&24\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 2 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&2&10&8&1&24\\& & \color{blue}{-4} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}-2&2&\color{orangered}{ 10 }&8&1&24\\& & \color{orangered}{-4} & & & \\ \hline &2&\color{orangered}{6}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 6 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&2&10&8&1&24\\& & -4& \color{blue}{-12} & & \\ \hline &2&\color{blue}{6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}-2&2&10&\color{orangered}{ 8 }&1&24\\& & -4& \color{orangered}{-12} & & \\ \hline &2&6&\color{orangered}{-4}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&2&10&8&1&24\\& & -4& -12& \color{blue}{8} & \\ \hline &2&6&\color{blue}{-4}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 8 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrr}-2&2&10&8&\color{orangered}{ 1 }&24\\& & -4& -12& \color{orangered}{8} & \\ \hline &2&6&-4&\color{orangered}{9}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 9 } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&2&10&8&1&24\\& & -4& -12& 8& \color{blue}{-18} \\ \hline &2&6&-4&\color{blue}{9}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 24 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}-2&2&10&8&1&\color{orangered}{ 24 }\\& & -4& -12& 8& \color{orangered}{-18} \\ \hline &\color{blue}{2}&\color{blue}{6}&\color{blue}{-4}&\color{blue}{9}&\color{orangered}{6} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+6x^{2}-4x+9 } $ with a remainder of $ \color{red}{ 6 } $.