Divide using synthetic division $$ ( 2x^{5}+4x^{2}+3 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-2&2&0&0&4&0&3\\& & -4& 8& -16& 24& \color{black}{-48} \\ \hline &\color{blue}{2}&\color{blue}{-4}&\color{blue}{8}&\color{blue}{-12}&\color{blue}{24}&\color{orangered}{-45} \end{array} $$The solution is:
$$ \frac{ 2x^{5}+4x^{2}+3 }{ x+2 } = \color{blue}{2x^{4}-4x^{3}+8x^{2}-12x+24} \color{red}{~-~} \frac{ \color{red}{ 45 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&2&0&0&4&0&3\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-2&\color{orangered}{ 2 }&0&0&4&0&3\\& & & & & & \\ \hline &\color{orangered}{2}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&2&0&0&4&0&3\\& & \color{blue}{-4} & & & & \\ \hline &\color{blue}{2}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrrr}-2&2&\color{orangered}{ 0 }&0&4&0&3\\& & \color{orangered}{-4} & & & & \\ \hline &2&\color{orangered}{-4}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&2&0&0&4&0&3\\& & -4& \color{blue}{8} & & & \\ \hline &2&\color{blue}{-4}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 8 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrrr}-2&2&0&\color{orangered}{ 0 }&4&0&3\\& & -4& \color{orangered}{8} & & & \\ \hline &2&-4&\color{orangered}{8}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 8 } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&2&0&0&4&0&3\\& & -4& 8& \color{blue}{-16} & & \\ \hline &2&-4&\color{blue}{8}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrrrr}-2&2&0&0&\color{orangered}{ 4 }&0&3\\& & -4& 8& \color{orangered}{-16} & & \\ \hline &2&-4&8&\color{orangered}{-12}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&2&0&0&4&0&3\\& & -4& 8& -16& \color{blue}{24} & \\ \hline &2&-4&8&\color{blue}{-12}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 24 } = \color{orangered}{ 24 } $
$$ \begin{array}{c|rrrrrr}-2&2&0&0&4&\color{orangered}{ 0 }&3\\& & -4& 8& -16& \color{orangered}{24} & \\ \hline &2&-4&8&-12&\color{orangered}{24}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 24 } = \color{blue}{ -48 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&2&0&0&4&0&3\\& & -4& 8& -16& 24& \color{blue}{-48} \\ \hline &2&-4&8&-12&\color{blue}{24}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -48 \right) } = \color{orangered}{ -45 } $
$$ \begin{array}{c|rrrrrr}-2&2&0&0&4&0&\color{orangered}{ 3 }\\& & -4& 8& -16& 24& \color{orangered}{-48} \\ \hline &\color{blue}{2}&\color{blue}{-4}&\color{blue}{8}&\color{blue}{-12}&\color{blue}{24}&\color{orangered}{-45} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{4}-4x^{3}+8x^{2}-12x+24 } $ with a remainder of $ \color{red}{ -45 } $.